第四讲 三角形的五心

$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{E}\htmlData{tutor-start=13,tutor-end=19}{\perp }\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{C}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=23}{C}\htmlData{tutor-start=23,tutor-end=24}{F}\htmlData{tutor-start=24,tutor-end=30}{\perp }\htmlData{tutor-start=30,tutor-end=31}{A}\htmlData{tutor-start=31,tutor-end=32}{B}$。因此 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆；循环考察另外两个顶点也有相同结构。

详细展开：由题意，$\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{C}$，$\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{C}$，$\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{F}\htmlData{tutor-start=2,tutor-end=8}{\perp }\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{B}$，垂足分别为 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{F}$，垂心为 $\htmlData{tutor-start=0,tutor-end=1}{H}$。于是 $\angle CEH=90^\circ$，$\angle CDH=90^\circ$，所以 $\angle CEH+\angle CDH=180^\circ$，即 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆（对角互补的四边形内接于圆）。同理，$\angle BFH=90^\circ$，$\angle BDH=90^\circ$，故 $\angle BFH+\angle BDH=180^\circ$，所以 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆。循环考察另外两个顶点 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{B}$、$\htmlData{tutor-start=0,tutor-end=1}{A}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}$，同样有：$\angle AEH=\angle AFH=90^\circ$ 推出 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{F}$ 四点共圆；$\angle AFH=\angle ADH=90^\circ$ 推出 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆；$\angle AEH=\angle ADH=90^\circ$ 推出 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 四点共圆。这些共圆结构为后续用圆周角相等建立角平分关系提供了基础。

由圆周角关系可得 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{D}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}$，$\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{E}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{F}$，$\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{F}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{E}$。锐角三角形的垂心位于垂足三角形内部。

详细展开：在圆 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{F}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 中，$\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}$ 与 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{F}$ 同对弧 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{F}$，故 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{B}\htmlData{tutor-start=20,tutor-end=21}{F}$。在圆 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{F}$（或等价地由 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{B}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{E}$ 共圆）中，$\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{F}$ 即 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{H}$，而 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{B}\htmlData{tutor-start=9,tutor-end=10}{H}$ 与 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{H}$ 同对弧 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{H}$（在圆 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{E}$ 中，由 $\angle BDH=\angle BEH=90^\circ$ 得 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{D}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{E}$ 共圆），故 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{B}\htmlData{tutor-start=20,tutor-end=21}{H}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{H}\htmlData{tutor-start=30,tutor-end=31}{B}\htmlData{tutor-start=31,tutor-end=32}{F}$。于是 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{H}$，再结合对称结构可得 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{D}\htmlData{tutor-start=20,tutor-end=21}{F}$，即 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{D}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}$。同理，在圆 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{F}$ 中 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{F}$，在圆 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 中 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{H}$，而 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{A}\htmlData{tutor-start=20,tutor-end=21}{H}$（因为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{F}$ 的夹角被 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 分成的两部分分别等于 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{H}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{F}$ 与 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{A}\htmlData{tutor-start=9,tutor-end=10}{H}$，且由 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{F}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{E}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{H}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{D}$ 的对称性可知它们相等），故 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{F}$，即 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{E}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{F}$。类似可得 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{F}$ 平分 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{E}$。由于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 为锐角三角形，垂心 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 位于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 内部。

三条内角平分线交于内心，所以 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 是三角形 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{F}$ 的内心，选择 C。

详细展开：由上一步已证 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{D}\htmlData{tutor-start=20,tutor-end=21}{F}$，$\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{D}\htmlData{tutor-start=8,tutor-end=9}{E}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{F}$，$\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{F}\htmlData{tutor-start=9,tutor-end=10}{H}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{H}\htmlData{tutor-start=19,tutor-end=20}{F}\htmlData{tutor-start=20,tutor-end=21}{D}$，即 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{D}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{H}\htmlData{tutor-start=4,tutor-end=5}{E}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{H}\htmlData{tutor-start=7,tutor-end=8}{F}$ 分别是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 三个内角 $\htmlData{tutor-start=0,tutor-end=7}{\angle }\htmlData{tutor-start=7,tutor-end=8}{E}\htmlData{tutor-start=8,tutor-end=9}{D}\htmlData{tutor-start=9,tutor-end=10}{F}\htmlData{tutor-start=10,tutor-end=11}{,}\htmlData{tutor-start=11,tutor-end=18}{\angle }\htmlData{tutor-start=18,tutor-end=19}{D}\htmlData{tutor-start=19,tutor-end=20}{E}\htmlData{tutor-start=20,tutor-end=21}{F}\htmlData{tutor-start=21,tutor-end=22}{,}\htmlData{tutor-start=22,tutor-end=29}{\angle }\htmlData{tutor-start=29,tutor-end=30}{D}\htmlData{tutor-start=30,tutor-end=31}{F}\htmlData{tutor-start=31,tutor-end=32}{E}$ 的角平分线。根据三角形内心的定义：三角形三条内角平分线交于一点，该点称为三角形的内心，记作 $\htmlData{tutor-start=0,tutor-end=1}{I}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{D}\htmlData{tutor-start=14,tutor-end=15}{E}\htmlData{tutor-start=15,tutor-end=16}{F}}$。由于 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 是这三条角平分线的公共交点，所以 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{I}_{\htmlData{tutor-start=5,tutor-end=15}{\triangle }\htmlData{tutor-start=15,tutor-end=16}{D}\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{F}}$，即 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{D}\htmlData{tutor-start=11,tutor-end=12}{E}\htmlData{tutor-start=12,tutor-end=13}{F}$ 的内心。对照选项，选 C。

取任意原点，用位置向量表示点。三角形重心等于三个顶点位置向量的平均值。

详细展开：首先，我们需要将几何图形中的点转化为代数对象以便计算。选取平面内任意一点 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 作为原点，用位置向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}}\htmlData{tutor-start=7,tutor-end=8}{,} \vec{\htmlData{tutor-start=14,tutor-end=15}{B}}\htmlData{tutor-start=16,tutor-end=17}{,} \vec{\htmlData{tutor-start=23,tutor-end=24}{C}}\htmlData{tutor-start=25,tutor-end=26}{,} \vec{\htmlData{tutor-start=32,tutor-end=33}{D}}$ 分别表示点 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{C}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{D}$。根据三角形重心的定义，三角形的重心是其三个顶点位置向量的算术平均值。对于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{D}$，其顶点为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{D}$，因此重心 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的位置向量为这三个向量和的三分之一。同理，对于 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{D}$，其顶点为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{C}\htmlData{tutor-start=4,tutor-end=5}{,} \htmlData{tutor-start=6,tutor-end=7}{D}$，重心 $\htmlData{tutor-start=0,tutor-end=1}{F}$ 的位置向量也是这三个向量和的三分之一。我们将这两个公式明确写出，作为后续推导的基础。

因为 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{D}$ 都在直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 上，$\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{F}$ 的表达式中 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}}$ 的系数同为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$。因此直线 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 上任一点相对于直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的仿射高度都是 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$。

详细展开：观察上一步得到的 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}}$ 和 $\vec{\htmlData{tutor-start=5,tutor-end=6}{F}}$ 的表达式，我们计算向量 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}}$ 以判断其与边 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$ 的关系。由向量减法定义，$\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}} \htmlData{tutor-start=9,tutor-end=10}{=} \vec{\htmlData{tutor-start=16,tutor-end=17}{F}} \htmlData{tutor-start=19,tutor-end=20}{-} \vec{\htmlData{tutor-start=26,tutor-end=27}{E}}$。代入具体表达式：$\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}} \htmlData{tutor-start=9,tutor-end=10}{=} \frac{\vec{\htmlData{tutor-start=22,tutor-end=23}{A}}\htmlData{tutor-start=24,tutor-end=25}{+}\vec{\htmlData{tutor-start=30,tutor-end=31}{C}}\htmlData{tutor-start=32,tutor-end=33}{+}\vec{\htmlData{tutor-start=38,tutor-end=39}{D}}}{\htmlData{tutor-start=42,tutor-end=43}{3}} \htmlData{tutor-start=45,tutor-end=46}{-} \frac{\vec{\htmlData{tutor-start=58,tutor-end=59}{A}}\htmlData{tutor-start=60,tutor-end=61}{+}\vec{\htmlData{tutor-start=66,tutor-end=67}{B}}\htmlData{tutor-start=68,tutor-end=69}{+}\vec{\htmlData{tutor-start=74,tutor-end=75}{D}}}{\htmlData{tutor-start=78,tutor-end=79}{3}}$。合并同类项，分子中的 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}}$ 和 $\vec{\htmlData{tutor-start=5,tutor-end=6}{D}}$ 相互抵消，剩下 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}} \htmlData{tutor-start=9,tutor-end=10}{=} \frac{\vec{\htmlData{tutor-start=22,tutor-end=23}{C}} \htmlData{tutor-start=25,tutor-end=26}{-} \vec{\htmlData{tutor-start=32,tutor-end=33}{B}}}{\htmlData{tutor-start=36,tutor-end=37}{3}}$。注意到 $\vec{\htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{C}} \htmlData{tutor-start=9,tutor-end=10}{=} \vec{\htmlData{tutor-start=16,tutor-end=17}{C}} \htmlData{tutor-start=19,tutor-end=20}{-} \vec{\htmlData{tutor-start=26,tutor-end=27}{B}}$，因此 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}} \htmlData{tutor-start=9,tutor-end=10}{=} \frac{\htmlData{tutor-start=17,tutor-end=18}{1}}{\htmlData{tutor-start=20,tutor-end=21}{3}}\vec{\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{C}}$。根据向量共线定理，若 $\vec{\htmlData{tutor-start=5,tutor-end=6}{u}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{k}\vec{\htmlData{tutor-start=16,tutor-end=17}{v}}$ 且 $\htmlData{tutor-start=0,tutor-end=1}{k} \neq \htmlData{tutor-start=7,tutor-end=8}{0}$，则两向量平行。这里 $\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}$，故 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}\htmlData{tutor-start=6,tutor-end=7}{F}} \htmlData{tutor-start=9,tutor-end=19}{\parallel }\vec{\htmlData{tutor-start=24,tutor-end=25}{B}\htmlData{tutor-start=25,tutor-end=26}{C}}$，即直线 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 平行于直线 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$。

设 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{t}\htmlData{tutor-start=6,tutor-end=7}{)}\htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{t}\htmlData{tutor-start=10,tutor-end=11}{D}$。交点 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 位于 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 上，故 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的系数 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}$，得到 $\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{/}\htmlData{tutor-start=4,tutor-end=5}{3}$。于是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{G}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{G}\htmlData{tutor-start=4,tutor-end=5}{D}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{:}\htmlData{tutor-start=8,tutor-end=9}{1}$。

详细展开：既然 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F} \htmlData{tutor-start=3,tutor-end=13}{\parallel }\htmlData{tutor-start=13,tutor-end=14}{B}\htmlData{tutor-start=14,tutor-end=15}{C}$，我们可以利用平行线截割定理或向量共线条件来确定 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 的位置。点 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 与 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 的交点。由于 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F}$ 分别是 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{D}$ 和 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{C}\htmlData{tutor-start=12,tutor-end=13}{D}$ 的重心，它们在“高度”方向（相对于底边 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$）上的位置是固定的。更严谨地，我们使用向量参数方程。设 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 上，则存在实数 $\htmlData{tutor-start=0,tutor-end=1}{t}$ 使得 $\vec{\htmlData{tutor-start=5,tutor-end=6}{G}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{)}\vec{\htmlData{tutor-start=20,tutor-end=21}{A}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{t}\vec{\htmlData{tutor-start=31,tutor-end=32}{D}}$。同时，$\htmlData{tutor-start=0,tutor-end=1}{G}$ 也在直线 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 上。考察 $\vec{\htmlData{tutor-start=5,tutor-end=6}{E}}$ 和 $\vec{\htmlData{tutor-start=5,tutor-end=6}{F}}$ 的表达式，它们都可以写成 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}}{\htmlData{tutor-start=9,tutor-end=10}{3}}\vec{\htmlData{tutor-start=16,tutor-end=17}{A}} \htmlData{tutor-start=19,tutor-end=20}{+} \frac{\htmlData{tutor-start=27,tutor-end=28}{1}}{\htmlData{tutor-start=30,tutor-end=31}{3}}\vec{\htmlData{tutor-start=37,tutor-end=38}{D}} \htmlData{tutor-start=40,tutor-end=41}{+} \text{\htmlData{tutor-start=48,tutor-end=49}{底}\htmlData{tutor-start=49,tutor-end=50}{边}\htmlData{tutor-start=50,tutor-end=51}{相}\htmlData{tutor-start=51,tutor-end=52}{关}\htmlData{tutor-start=52,tutor-end=53}{项}}$ 的形式。具体来说，直线 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 上的任意点 $\htmlData{tutor-start=0,tutor-end=1}{P}$ 可以表示为 $\htmlData{tutor-start=0,tutor-end=8}{\lambda }\vec{\htmlData{tutor-start=13,tutor-end=14}{E}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=28}{\lambda}\htmlData{tutor-start=28,tutor-end=29}{)}\vec{\htmlData{tutor-start=34,tutor-end=35}{F}}$。代入 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{F}$ 的表达式并整理，发现 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}}$ 和 $\vec{\htmlData{tutor-start=5,tutor-end=6}{D}}$ 的系数始终为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$。因为 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 在 $\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{F}$ 上，所以 $\vec{\htmlData{tutor-start=5,tutor-end=6}{G}}$ 分解式中 $\vec{\htmlData{tutor-start=5,tutor-end=6}{A}}$ 的系数必须是 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{/}\htmlData{tutor-start=2,tutor-end=3}{3}$。对比 $\vec{\htmlData{tutor-start=5,tutor-end=6}{G}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{)}\vec{\htmlData{tutor-start=20,tutor-end=21}{A}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{t}\vec{\htmlData{tutor-start=31,tutor-end=32}{D}}$，可得 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{t} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{/}\htmlData{tutor-start=8,tutor-end=9}{3}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{3}$。这意味着 $\vec{\htmlData{tutor-start=5,tutor-end=6}{G}} \htmlData{tutor-start=8,tutor-end=9}{=} \frac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{3}}\vec{\htmlData{tutor-start=26,tutor-end=27}{A}} \htmlData{tutor-start=29,tutor-end=30}{+} \frac{\htmlData{tutor-start=37,tutor-end=38}{2}}{\htmlData{tutor-start=40,tutor-end=41}{3}}\vec{\htmlData{tutor-start=47,tutor-end=48}{D}}$。根据定比分点公式，$\htmlData{tutor-start=0,tutor-end=1}{G}$ 分 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D}$ 的比为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{G}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{G}\htmlData{tutor-start=4,tutor-end=5}{D} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{:}\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{t}\htmlData{tutor-start=14,tutor-end=15}{)} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{/}\htmlData{tutor-start=20,tutor-end=21}{3} \htmlData{tutor-start=22,tutor-end=23}{:} \htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{/}\htmlData{tutor-start=26,tutor-end=27}{3} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{:}\htmlData{tutor-start=32,tutor-end=33}{1}$。题目求 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{G}\htmlData{tutor-start=2,tutor-end=3}{:}\htmlData{tutor-start=3,tutor-end=4}{G}\htmlData{tutor-start=4,tutor-end=5}{A}$，即 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{:}\htmlData{tutor-start=2,tutor-end=3}{2}$。

以 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 为原点，记 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}}\htmlData{tutor-start=10,tutor-end=11}{=}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{G}\htmlData{tutor-start=28,tutor-end=29}{A}}\htmlData{tutor-start=30,tutor-end=31}{,}\mathbf{\htmlData{tutor-start=39,tutor-end=40}{y}}\htmlData{tutor-start=41,tutor-end=42}{=}\overrightarrow{\htmlData{tutor-start=58,tutor-end=59}{G}\htmlData{tutor-start=59,tutor-end=60}{B}}\htmlData{tutor-start=61,tutor-end=62}{,}\mathbf{\htmlData{tutor-start=70,tutor-end=71}{z}}\htmlData{tutor-start=72,tutor-end=73}{=}\overrightarrow{\htmlData{tutor-start=89,tutor-end=90}{G}\htmlData{tutor-start=90,tutor-end=91}{C}}$。重心关系给出 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}}\htmlData{tutor-start=10,tutor-end=11}{+}\mathbf{\htmlData{tutor-start=19,tutor-end=20}{y}}\htmlData{tutor-start=21,tutor-end=22}{+}\mathbf{\htmlData{tutor-start=30,tutor-end=31}{z}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{0}$。

详细展开：首先观察题目给出的几何结构：三角形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的重心为 $\htmlData{tutor-start=0,tutor-end=1}{G}$，且已知从重心到三个顶点的距离分别为 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{3}}$，$\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\sqrt{\htmlData{tutor-start=10,tutor-end=11}{2}}$，$\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}$。为了利用向量工具处理长度和角度关系，我们建立以重心 $\htmlData{tutor-start=0,tutor-end=1}{G}$ 为原点的平面直角坐标系或直接使用自由向量。记向量 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=12}{=} \overrightarrow{\htmlData{tutor-start=29,tutor-end=30}{G}\htmlData{tutor-start=30,tutor-end=31}{A}}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{y}} \htmlData{tutor-start=11,tutor-end=12}{=} \overrightarrow{\htmlData{tutor-start=29,tutor-end=30}{G}\htmlData{tutor-start=30,tutor-end=31}{B}}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{z}} \htmlData{tutor-start=11,tutor-end=12}{=} \overrightarrow{\htmlData{tutor-start=29,tutor-end=30}{G}\htmlData{tutor-start=30,tutor-end=31}{C}}$。根据重心的向量性质，对于任意三角形，其重心到三个顶点的向量之和为零向量，即 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=12}{+} \mathbf{\htmlData{tutor-start=21,tutor-end=22}{y}} \htmlData{tutor-start=24,tutor-end=25}{+} \mathbf{\htmlData{tutor-start=34,tutor-end=35}{z}} \htmlData{tutor-start=37,tutor-end=38}{=} \mathbf{\htmlData{tutor-start=47,tutor-end=48}{0}}$。同时，将已知的线段长度转化为向量的模长平方：$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{G}\htmlData{tutor-start=20,tutor-end=21}{A}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{2}\sqrt{\htmlData{tutor-start=36,tutor-end=37}{3}}\htmlData{tutor-start=38,tutor-end=39}{)}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{2}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{y}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{G}\htmlData{tutor-start=20,tutor-end=21}{B}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{2}\sqrt{\htmlData{tutor-start=36,tutor-end=37}{2}}\htmlData{tutor-start=38,tutor-end=39}{)}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{8}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{z}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{G}\htmlData{tutor-start=20,tutor-end=21}{C}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{2}^{\htmlData{tutor-start=31,tutor-end=32}{2}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{4}$。这些条件构成了后续计算的基础方程组。

由 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{z}}\htmlData{tutor-start=10,tutor-end=11}{=}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{(}\mathbf{\htmlData{tutor-start=21,tutor-end=22}{x}}\htmlData{tutor-start=23,tutor-end=24}{+}\mathbf{\htmlData{tutor-start=32,tutor-end=33}{y}}\htmlData{tutor-start=34,tutor-end=35}{)}$，两边取模平方，代入三个已知长度。

详细展开：我们的目标是求出向量之间的夹角信息，这可以通过数量积体现。由步骤 S1 中的关系式 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=12}{+} \mathbf{\htmlData{tutor-start=21,tutor-end=22}{y}} \htmlData{tutor-start=24,tutor-end=25}{+} \mathbf{\htmlData{tutor-start=34,tutor-end=35}{z}} \htmlData{tutor-start=37,tutor-end=38}{=} \mathbf{\htmlData{tutor-start=47,tutor-end=48}{0}}$，我们可以移项得到 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{z}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{(}\mathbf{\htmlData{tutor-start=23,tutor-end=24}{x}} \htmlData{tutor-start=26,tutor-end=27}{+} \mathbf{\htmlData{tutor-start=36,tutor-end=37}{y}}\htmlData{tutor-start=38,tutor-end=39}{)}$。为了利用已知的模长，对等式两边同时取模的平方。左边为 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{z}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}}$，右边为 $\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{(}\mathbf{\htmlData{tutor-start=11,tutor-end=12}{x}} \htmlData{tutor-start=14,tutor-end=15}{+} \mathbf{\htmlData{tutor-start=24,tutor-end=25}{y}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{|}^{\htmlData{tutor-start=30,tutor-end=31}{2}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{|}\mathbf{\htmlData{tutor-start=44,tutor-end=45}{x}} \htmlData{tutor-start=47,tutor-end=48}{+} \mathbf{\htmlData{tutor-start=57,tutor-end=58}{y}}\htmlData{tutor-start=59,tutor-end=60}{|}^{\htmlData{tutor-start=62,tutor-end=63}{2}}$。根据向量模长平方公式 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{a}} \htmlData{tutor-start=12,tutor-end=13}{+} \mathbf{\htmlData{tutor-start=22,tutor-end=23}{b}}\htmlData{tutor-start=24,tutor-end=25}{|}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{|}\mathbf{\htmlData{tutor-start=41,tutor-end=42}{a}}\htmlData{tutor-start=43,tutor-end=44}{|}^{\htmlData{tutor-start=46,tutor-end=47}{2}} \htmlData{tutor-start=49,tutor-end=50}{+} \htmlData{tutor-start=51,tutor-end=52}{|}\mathbf{\htmlData{tutor-start=60,tutor-end=61}{b}}\htmlData{tutor-start=62,tutor-end=63}{|}^{\htmlData{tutor-start=65,tutor-end=66}{2}} \htmlData{tutor-start=68,tutor-end=69}{+} \htmlData{tutor-start=70,tutor-end=71}{2}\mathbf{\htmlData{tutor-start=79,tutor-end=80}{a}} \htmlData{tutor-start=82,tutor-end=88}{\cdot }\mathbf{\htmlData{tutor-start=96,tutor-end=97}{b}}$，展开右边得到 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{|}\mathbf{\htmlData{tutor-start=28,tutor-end=29}{y}}\htmlData{tutor-start=30,tutor-end=31}{|}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{+} \htmlData{tutor-start=38,tutor-end=39}{2}\mathbf{\htmlData{tutor-start=47,tutor-end=48}{x}} \htmlData{tutor-start=50,tutor-end=56}{\cdot }\mathbf{\htmlData{tutor-start=64,tutor-end=65}{y}}$。代入已知数值：$\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{2} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{8} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\mathbf{\htmlData{tutor-start=22,tutor-end=23}{x}} \htmlData{tutor-start=25,tutor-end=31}{\cdot }\mathbf{\htmlData{tutor-start=39,tutor-end=40}{y}}$。整理方程：$\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{0} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{2}\mathbf{\htmlData{tutor-start=18,tutor-end=19}{x}} \htmlData{tutor-start=21,tutor-end=27}{\cdot }\mathbf{\htmlData{tutor-start=35,tutor-end=36}{y}}$，移项得 $\htmlData{tutor-start=0,tutor-end=1}{2}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=18}{\cdot }\mathbf{\htmlData{tutor-start=26,tutor-end=27}{y}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{4} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{0} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{6}$，解得 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{y}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{8}$。这一步成功求出了向量 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{G}\htmlData{tutor-start=17,tutor-end=18}{A}}$ 与 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{G}\htmlData{tutor-start=17,tutor-end=18}{B}}$ 的数量积。

由 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=17}{\times}\mathbf{\htmlData{tutor-start=25,tutor-end=26}{y}}\htmlData{tutor-start=27,tutor-end=28}{|}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{|}\mathbf{\htmlData{tutor-start=42,tutor-end=43}{x}}\htmlData{tutor-start=44,tutor-end=45}{|}^{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{|}\mathbf{\htmlData{tutor-start=58,tutor-end=59}{y}}\htmlData{tutor-start=60,tutor-end=61}{|}^{\htmlData{tutor-start=63,tutor-end=64}{2}}\htmlData{tutor-start=65,tutor-end=66}{-}\htmlData{tutor-start=66,tutor-end=67}{(}\mathbf{\htmlData{tutor-start=75,tutor-end=76}{x}}\htmlData{tutor-start=77,tutor-end=82}{\cdot}\mathbf{\htmlData{tutor-start=90,tutor-end=91}{y}}\htmlData{tutor-start=92,tutor-end=93}{)}^{\htmlData{tutor-start=95,tutor-end=96}{2}}$ 得到由 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{G}\htmlData{tutor-start=4,tutor-end=5}{B}$ 张成的平行四边形面积。

详细展开：有了数量积 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{y}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{8}$ 和模长 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{=}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{|}\mathbf{\htmlData{tutor-start=33,tutor-end=34}{y}}\htmlData{tutor-start=35,tutor-end=36}{|}\htmlData{tutor-start=36,tutor-end=37}{=}\sqrt{\htmlData{tutor-start=43,tutor-end=44}{8}}$，我们可以计算由向量 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{x}}$ 和 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{y}}$ 张成的平行四边形的面积，即叉积的模长 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=19}{\times }\mathbf{\htmlData{tutor-start=27,tutor-end=28}{y}}\htmlData{tutor-start=29,tutor-end=30}{|}$。在二维平面中，利用恒等式 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=19}{\times }\mathbf{\htmlData{tutor-start=27,tutor-end=28}{y}}\htmlData{tutor-start=29,tutor-end=30}{|}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{|}\mathbf{\htmlData{tutor-start=46,tutor-end=47}{x}}\htmlData{tutor-start=48,tutor-end=49}{|}^{\htmlData{tutor-start=51,tutor-end=52}{2}} \htmlData{tutor-start=54,tutor-end=55}{|}\mathbf{\htmlData{tutor-start=63,tutor-end=64}{y}}\htmlData{tutor-start=65,tutor-end=66}{|}^{\htmlData{tutor-start=68,tutor-end=69}{2}} \htmlData{tutor-start=71,tutor-end=72}{-} \htmlData{tutor-start=73,tutor-end=74}{(}\mathbf{\htmlData{tutor-start=82,tutor-end=83}{x}} \htmlData{tutor-start=85,tutor-end=91}{\cdot }\mathbf{\htmlData{tutor-start=99,tutor-end=100}{y}}\htmlData{tutor-start=101,tutor-end=102}{)}^{\htmlData{tutor-start=104,tutor-end=105}{2}}$（这是拉格朗日恒等式的二维形式，也等价于 $\sin^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=15}{\theta }\htmlData{tutor-start=15,tutor-end=16}{+} \cos^{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=32}{\theta }\htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{1}$ 的向量表达）。代入数值：$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=19}{\times }\mathbf{\htmlData{tutor-start=27,tutor-end=28}{y}}\htmlData{tutor-start=29,tutor-end=30}{|}^{\htmlData{tutor-start=32,tutor-end=33}{2}} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{2} \htmlData{tutor-start=40,tutor-end=46}{\cdot }\htmlData{tutor-start=46,tutor-end=47}{8} \htmlData{tutor-start=48,tutor-end=49}{-} \htmlData{tutor-start=50,tutor-end=51}{(}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{8}\htmlData{tutor-start=53,tutor-end=54}{)}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{9}\htmlData{tutor-start=62,tutor-end=63}{6} \htmlData{tutor-start=64,tutor-end=65}{-} \htmlData{tutor-start=66,tutor-end=67}{6}\htmlData{tutor-start=67,tutor-end=68}{4} \htmlData{tutor-start=69,tutor-end=70}{=} \htmlData{tutor-start=71,tutor-end=72}{3}\htmlData{tutor-start=72,tutor-end=73}{2}$。因此，$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}} \htmlData{tutor-start=12,tutor-end=19}{\times }\mathbf{\htmlData{tutor-start=27,tutor-end=28}{y}}\htmlData{tutor-start=29,tutor-end=30}{|} \htmlData{tutor-start=31,tutor-end=32}{=} \sqrt{\htmlData{tutor-start=39,tutor-end=40}{3}\htmlData{tutor-start=40,tutor-end=41}{2}} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{4}\sqrt{\htmlData{tutor-start=52,tutor-end=53}{2}}$。这个值代表了以 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{A}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{B}$ 为邻边的平行四边形面积，也是三角形 $\htmlData{tutor-start=0,tutor-end=1}{G}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{B}$ 面积的两倍。

$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}}\htmlData{tutor-start=19,tutor-end=20}{=}\mathbf{\htmlData{tutor-start=28,tutor-end=29}{y}}\htmlData{tutor-start=30,tutor-end=31}{-}\mathbf{\htmlData{tutor-start=39,tutor-end=40}{x}}$，$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}}\htmlData{tutor-start=19,tutor-end=20}{=}\mathbf{\htmlData{tutor-start=28,tutor-end=29}{z}}\htmlData{tutor-start=30,tutor-end=31}{-}\mathbf{\htmlData{tutor-start=39,tutor-end=40}{x}}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{2}\mathbf{\htmlData{tutor-start=52,tutor-end=53}{x}}\htmlData{tutor-start=54,tutor-end=55}{-}\mathbf{\htmlData{tutor-start=63,tutor-end=64}{y}}$，其叉积为 $\htmlData{tutor-start=0,tutor-end=1}{3}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{x}}\htmlData{tutor-start=11,tutor-end=17}{\times}\mathbf{\htmlData{tutor-start=25,tutor-end=26}{y}}$。

详细展开：最后，我们将重心向量的叉积转换为原三角形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的面积。三角形面积公式为 $\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{|}\overrightarrow{\htmlData{tutor-start=49,tutor-end=50}{A}\htmlData{tutor-start=50,tutor-end=51}{B}} \htmlData{tutor-start=53,tutor-end=60}{\times }\overrightarrow{\htmlData{tutor-start=76,tutor-end=77}{A}\htmlData{tutor-start=77,tutor-end=78}{C}}\htmlData{tutor-start=79,tutor-end=80}{|}$。用重心向量表示边向量：$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}} \htmlData{tutor-start=20,tutor-end=21}{=} \mathbf{\htmlData{tutor-start=30,tutor-end=31}{y}} \htmlData{tutor-start=33,tutor-end=34}{-} \mathbf{\htmlData{tutor-start=43,tutor-end=44}{x}}$，$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=21}{=} \mathbf{\htmlData{tutor-start=30,tutor-end=31}{z}} \htmlData{tutor-start=33,tutor-end=34}{-} \mathbf{\htmlData{tutor-start=43,tutor-end=44}{x}}$。利用 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{z}} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{-}\mathbf{\htmlData{tutor-start=22,tutor-end=23}{x}} \htmlData{tutor-start=25,tutor-end=26}{-} \mathbf{\htmlData{tutor-start=35,tutor-end=36}{y}}$，代入得 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{-}\mathbf{\htmlData{tutor-start=32,tutor-end=33}{x}} \htmlData{tutor-start=35,tutor-end=36}{-} \mathbf{\htmlData{tutor-start=45,tutor-end=46}{y}}\htmlData{tutor-start=47,tutor-end=48}{)} \htmlData{tutor-start=49,tutor-end=50}{-} \mathbf{\htmlData{tutor-start=59,tutor-end=60}{x}} \htmlData{tutor-start=62,tutor-end=63}{=} \htmlData{tutor-start=64,tutor-end=65}{-}\htmlData{tutor-start=65,tutor-end=66}{2}\mathbf{\htmlData{tutor-start=74,tutor-end=75}{x}} \htmlData{tutor-start=77,tutor-end=78}{-} \mathbf{\htmlData{tutor-start=87,tutor-end=88}{y}}$。计算叉积：$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{B}} \htmlData{tutor-start=20,tutor-end=27}{\times }\overrightarrow{\htmlData{tutor-start=43,tutor-end=44}{A}\htmlData{tutor-start=44,tutor-end=45}{C}} \htmlData{tutor-start=47,tutor-end=48}{=} \htmlData{tutor-start=49,tutor-end=50}{(}\mathbf{\htmlData{tutor-start=58,tutor-end=59}{y}} \htmlData{tutor-start=61,tutor-end=62}{-} \mathbf{\htmlData{tutor-start=71,tutor-end=72}{x}}\htmlData{tutor-start=73,tutor-end=74}{)} \htmlData{tutor-start=75,tutor-end=82}{\times }\htmlData{tutor-start=82,tutor-end=83}{(}\htmlData{tutor-start=83,tutor-end=84}{-}\htmlData{tutor-start=84,tutor-end=85}{2}\mathbf{\htmlData{tutor-start=93,tutor-end=94}{x}} \htmlData{tutor-start=96,tutor-end=97}{-} \mathbf{\htmlData{tutor-start=106,tutor-end=107}{y}}\htmlData{tutor-start=108,tutor-end=109}{)}$。展开利用分配律和 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{a}} \htmlData{tutor-start=11,tutor-end=18}{\times }\mathbf{\htmlData{tutor-start=26,tutor-end=27}{a}} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{,} \mathbf{\htmlData{tutor-start=42,tutor-end=43}{y}} \htmlData{tutor-start=45,tutor-end=52}{\times }\mathbf{\htmlData{tutor-start=60,tutor-end=61}{x}} \htmlData{tutor-start=63,tutor-end=64}{=} \htmlData{tutor-start=65,tutor-end=66}{-}\mathbf{\htmlData{tutor-start=74,tutor-end=75}{x}} \htmlData{tutor-start=77,tutor-end=84}{\times }\mathbf{\htmlData{tutor-start=92,tutor-end=93}{y}}$：原式 $\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{(}\mathbf{\htmlData{tutor-start=13,tutor-end=14}{y}} \htmlData{tutor-start=16,tutor-end=23}{\times }\mathbf{\htmlData{tutor-start=31,tutor-end=32}{x}}\htmlData{tutor-start=33,tutor-end=34}{)} \htmlData{tutor-start=35,tutor-end=36}{-} \htmlData{tutor-start=37,tutor-end=38}{(}\mathbf{\htmlData{tutor-start=46,tutor-end=47}{y}} \htmlData{tutor-start=49,tutor-end=56}{\times }\mathbf{\htmlData{tutor-start=64,tutor-end=65}{y}}\htmlData{tutor-start=66,tutor-end=67}{)} \htmlData{tutor-start=68,tutor-end=69}{+} \htmlData{tutor-start=70,tutor-end=71}{2}\htmlData{tutor-start=71,tutor-end=72}{(}\mathbf{\htmlData{tutor-start=80,tutor-end=81}{x}} \htmlData{tutor-start=83,tutor-end=90}{\times }\mathbf{\htmlData{tutor-start=98,tutor-end=99}{x}}\htmlData{tutor-start=100,tutor-end=101}{)} \htmlData{tutor-start=102,tutor-end=103}{+} \htmlData{tutor-start=104,tutor-end=105}{(}\mathbf{\htmlData{tutor-start=113,tutor-end=114}{x}} \htmlData{tutor-start=116,tutor-end=123}{\times }\mathbf{\htmlData{tutor-start=131,tutor-end=132}{y}}\htmlData{tutor-start=133,tutor-end=134}{)} \htmlData{tutor-start=135,tutor-end=136}{=} \htmlData{tutor-start=137,tutor-end=138}{2}\htmlData{tutor-start=138,tutor-end=139}{(}\mathbf{\htmlData{tutor-start=147,tutor-end=148}{x}} \htmlData{tutor-start=150,tutor-end=157}{\times }\mathbf{\htmlData{tutor-start=165,tutor-end=166}{y}}\htmlData{tutor-start=167,tutor-end=168}{)} \htmlData{tutor-start=169,tutor-end=170}{-} \htmlData{tutor-start=171,tutor-end=172}{0} \htmlData{tutor-start=173,tutor-end=174}{+} \htmlData{tutor-start=175,tutor-end=176}{0} \htmlData{tutor-start=177,tutor-end=178}{+} \htmlData{tutor-start=179,tutor-end=180}{(}\mathbf{\htmlData{tutor-start=188,tutor-end=189}{x}} \htmlData{tutor-start=191,tutor-end=198}{\times }\mathbf{\htmlData{tutor-start=206,tutor-end=207}{y}}\htmlData{tutor-start=208,tutor-end=209}{)} \htmlData{tutor-start=210,tutor-end=211}{=} \htmlData{tutor-start=212,tutor-end=213}{3}\htmlData{tutor-start=213,tutor-end=214}{(}\mathbf{\htmlData{tutor-start=222,tutor-end=223}{x}} \htmlData{tutor-start=225,tutor-end=232}{\times }\mathbf{\htmlData{tutor-start=240,tutor-end=241}{y}}\htmlData{tutor-start=242,tutor-end=243}{)}$。因此，$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{|}\htmlData{tutor-start=33,tutor-end=34}{3}\htmlData{tutor-start=34,tutor-end=35}{(}\mathbf{\htmlData{tutor-start=43,tutor-end=44}{x}} \htmlData{tutor-start=46,tutor-end=53}{\times }\mathbf{\htmlData{tutor-start=61,tutor-end=62}{y}}\htmlData{tutor-start=63,tutor-end=64}{)}\htmlData{tutor-start=64,tutor-end=65}{|} \htmlData{tutor-start=66,tutor-end=67}{=} \frac{\htmlData{tutor-start=74,tutor-end=75}{3}}{\htmlData{tutor-start=77,tutor-end=78}{2}} \htmlData{tutor-start=80,tutor-end=81}{|}\mathbf{\htmlData{tutor-start=89,tutor-end=90}{x}} \htmlData{tutor-start=92,tutor-end=99}{\times }\mathbf{\htmlData{tutor-start=107,tutor-end=108}{y}}\htmlData{tutor-start=109,tutor-end=110}{|}$。代入 S3 的结果：$\htmlData{tutor-start=0,tutor-end=1}{S}_{\htmlData{tutor-start=3,tutor-end=13}{\triangle }\htmlData{tutor-start=13,tutor-end=14}{A}\htmlData{tutor-start=14,tutor-end=15}{B}\htmlData{tutor-start=15,tutor-end=16}{C}} \htmlData{tutor-start=18,tutor-end=19}{=} \frac{\htmlData{tutor-start=26,tutor-end=27}{3}}{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=38}{\cdot }\htmlData{tutor-start=38,tutor-end=39}{4}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{2}} \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{6}\sqrt{\htmlData{tutor-start=57,tutor-end=58}{2}}$。

设外接圆半径为 $\htmlData{tutor-start=0,tutor-end=1}{R}$。垂心到顶点的距离满足 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{R}\htmlData{tutor-start=5,tutor-end=6}{|}\cos \htmlData{tutor-start=11,tutor-end=12}{A}\htmlData{tutor-start=12,tutor-end=13}{|}$，而扩充正弦定理给出 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{2}\htmlData{tutor-start=4,tutor-end=5}{R}\sin \htmlData{tutor-start=10,tutor-end=11}{A}$。

详细展开：首先，我们需要将题目中给出的几何长度条件转化为关于角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的三角函数表达式。设 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 的外接圆半径为 $\htmlData{tutor-start=0,tutor-end=1}{R}$。根据三角形垂心的性质，顶点 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 到垂心 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 的距离公式为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{R} \htmlData{tutor-start=8,tutor-end=9}{|}\\\htmlData{tutor-start=11,tutor-end=12}{c}\htmlData{tutor-start=12,tutor-end=13}{o}\htmlData{tutor-start=13,tutor-end=14}{s} \htmlData{tutor-start=15,tutor-end=16}{A}\htmlData{tutor-start=16,tutor-end=17}{|}$。这里使用绝对值是因为当 $\htmlData{tutor-start=0,tutor-end=10}{\triangle }\htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{B}\htmlData{tutor-start=12,tutor-end=13}{C}$ 为钝角三角形时，$\\\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{s} \htmlData{tutor-start=6,tutor-end=7}{A}$ 为负值，但距离必须为正。同时，根据正弦定理，边长 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}$（即角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 的对边 $\htmlData{tutor-start=0,tutor-end=1}{a}$）可以表示为 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{R} \\\htmlData{tutor-start=10,tutor-end=11}{s}\htmlData{tutor-start=11,tutor-end=12}{i}\htmlData{tutor-start=12,tutor-end=13}{n} \htmlData{tutor-start=14,tutor-end=15}{A}$。由于 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是三角形的内角，其范围在 (0^\\circ, 180^\\circ) 之间，因此 $\\\htmlData{tutor-start=2,tutor-end=3}{s}\htmlData{tutor-start=3,tutor-end=4}{i}\htmlData{tutor-start=4,tutor-end=5}{n} \htmlData{tutor-start=6,tutor-end=7}{A}$ 恒为正数，无需加绝对值。这一步建立了已知等式 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}$ 与角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 之间的直接联系。

由 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{B}\htmlData{tutor-start=4,tutor-end=5}{C}$ 且 $0^\circ<A<180^\circ$，得到 $\htmlData{tutor-start=0,tutor-end=1}{|}\cos \htmlData{tutor-start=6,tutor-end=7}{A}\htmlData{tutor-start=7,tutor-end=8}{|}\htmlData{tutor-start=8,tutor-end=9}{=}\sin \htmlData{tutor-start=14,tutor-end=15}{A}$。锐角和钝角各有一个解。

详细展开：根据题设条件 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{B}\htmlData{tutor-start=6,tutor-end=7}{C}$，我们将上一步得到的表达式代入该等式：$\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{R} \htmlData{tutor-start=3,tutor-end=4}{|}\\\htmlData{tutor-start=6,tutor-end=7}{c}\htmlData{tutor-start=7,tutor-end=8}{o}\htmlData{tutor-start=8,tutor-end=9}{s} \htmlData{tutor-start=10,tutor-end=11}{A}\htmlData{tutor-start=11,tutor-end=12}{|} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{R} \\\htmlData{tutor-start=20,tutor-end=21}{s}\htmlData{tutor-start=21,tutor-end=22}{i}\htmlData{tutor-start=22,tutor-end=23}{n} \htmlData{tutor-start=24,tutor-end=25}{A}$。由于三角形存在，外接圆半径 $\htmlData{tutor-start=0,tutor-end=1}{R} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}$，我们可以从等式两边同时消去非零常数 $\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{R}$，得到简化后的三角方程 $\htmlData{tutor-start=0,tutor-end=1}{|}\\\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{o}\htmlData{tutor-start=5,tutor-end=6}{s} \htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{|} \htmlData{tutor-start=10,tutor-end=11}{=} \\\htmlData{tutor-start=14,tutor-end=15}{s}\htmlData{tutor-start=15,tutor-end=16}{i}\htmlData{tutor-start=16,tutor-end=17}{n} \htmlData{tutor-start=18,tutor-end=19}{A}$。接下来需要求解这个方程。因为 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 是三角形内角，所以 0^\\circ < A < 180^\\circ，在此范围内 $\\\htmlData{tutor-start=2,tutor-end=3}{s}\htmlData{tutor-start=3,tutor-end=4}{i}\htmlData{tutor-start=4,tutor-end=5}{n} \htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{>} \htmlData{tutor-start=10,tutor-end=11}{0}$。我们将讨论分为两种情况：1. 当 0^\\circ < A \\le 90^\\circ 时，$\\\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{s} \htmlData{tutor-start=6,tutor-end=7}{A} \\\htmlData{tutor-start=10,tutor-end=11}{g}\htmlData{tutor-start=11,tutor-end=12}{e} \htmlData{tutor-start=13,tutor-end=14}{0}$，此时 $\htmlData{tutor-start=0,tutor-end=1}{|}\\\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{o}\htmlData{tutor-start=5,tutor-end=6}{s} \htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{|} \htmlData{tutor-start=10,tutor-end=11}{=} \\\htmlData{tutor-start=14,tutor-end=15}{c}\htmlData{tutor-start=15,tutor-end=16}{o}\htmlData{tutor-start=16,tutor-end=17}{s} \htmlData{tutor-start=18,tutor-end=19}{A}$，方程变为 $\\\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{s} \htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{=} \\\htmlData{tutor-start=12,tutor-end=13}{s}\htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{n} \htmlData{tutor-start=16,tutor-end=17}{A}$，即 $\\\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{n} \htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{1}$，解得 A = 45^\\circ。2. 当 90^\\circ < A < 180^\\circ 时，$\\\htmlData{tutor-start=2,tutor-end=3}{c}\htmlData{tutor-start=3,tutor-end=4}{o}\htmlData{tutor-start=4,tutor-end=5}{s} \htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{<} \htmlData{tutor-start=10,tutor-end=11}{0}$，此时 $\htmlData{tutor-start=0,tutor-end=1}{|}\\\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{o}\htmlData{tutor-start=5,tutor-end=6}{s} \htmlData{tutor-start=7,tutor-end=8}{A}\htmlData{tutor-start=8,tutor-end=9}{|} \htmlData{tutor-start=10,tutor-end=11}{=} \htmlData{tutor-start=12,tutor-end=13}{-}\\\htmlData{tutor-start=15,tutor-end=16}{c}\htmlData{tutor-start=16,tutor-end=17}{o}\htmlData{tutor-start=17,tutor-end=18}{s} \htmlData{tutor-start=19,tutor-end=20}{A}$，方程变为 $\htmlData{tutor-start=0,tutor-end=1}{-}\\\htmlData{tutor-start=3,tutor-end=4}{c}\htmlData{tutor-start=4,tutor-end=5}{o}\htmlData{tutor-start=5,tutor-end=6}{s} \htmlData{tutor-start=7,tutor-end=8}{A} \htmlData{tutor-start=9,tutor-end=10}{=} \\\htmlData{tutor-start=13,tutor-end=14}{s}\htmlData{tutor-start=14,tutor-end=15}{i}\htmlData{tutor-start=15,tutor-end=16}{n} \htmlData{tutor-start=17,tutor-end=18}{A}$，即 $\\\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{a}\htmlData{tutor-start=4,tutor-end=5}{n} \htmlData{tutor-start=6,tutor-end=7}{A} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{1}$，在第二象限解得 A = 135^\\circ。

两个角都符合原条件，因此选择 D。

详细展开：回顾题目要求，我们需要确定角 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{C}$ 的所有可能度数，并从选项中选择正确答案。经过前两步的推导，我们得出角 $\htmlData{tutor-start=0,tutor-end=1}{A}$ 可以是 45^\\circ（对应锐角三角形情形）或者 135^\\circ（对应钝角三角形情形）。这两个解均满足三角形内角和定理及题设的几何约束。对比给出的选项：A 仅包含 45^\\circ，不完整；B 仅包含 30^\\circ，错误；C 包含 30^\\circ 或 150^\\circ，数值错误；D 包含 45^\\circ 或 135^\\circ，完全覆盖了我们的计算结果。因此，正确选项是 D。

取 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 为原点，令 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}}\htmlData{tutor-start=10,tutor-end=11}{=}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{A}}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{v}}\htmlData{tutor-start=10,tutor-end=11}{=}\overrightarrow{\htmlData{tutor-start=27,tutor-end=28}{B}\htmlData{tutor-start=28,tutor-end=29}{C}}$。则 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=12}{|}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{0}$，$\htmlData{tutor-start=0,tutor-end=1}{D}$ 的位置向量为 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}}\htmlData{tutor-start=10,tutor-end=11}{+}\mathbf{\htmlData{tutor-start=19,tutor-end=20}{v}}$，中点 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 的位置向量为 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}}\htmlData{tutor-start=10,tutor-end=11}{/}\htmlData{tutor-start=11,tutor-end=12}{2}$。

详细展开：为了利用向量运算处理长度和角度关系，我们首先建立向量基底。选取点 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 作为坐标原点（参考点），定义两个基向量：令 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}} \htmlData{tutor-start=11,tutor-end=12}{=} \overrightarrow{\htmlData{tutor-start=29,tutor-end=30}{B}\htmlData{tutor-start=30,tutor-end=31}{A}}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{v}} \htmlData{tutor-start=11,tutor-end=12}{=} \overrightarrow{\htmlData{tutor-start=29,tutor-end=30}{B}\htmlData{tutor-start=30,tutor-end=31}{C}}$。根据题意，平行四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 的边 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 长度为 10，因此 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=12}{|} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{0}$。由于 $\htmlData{tutor-start=0,tutor-end=1}{E}$ 是 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}$ 的中点，且 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{A}} \htmlData{tutor-start=20,tutor-end=21}{=} \mathbf{\htmlData{tutor-start=30,tutor-end=31}{u}}$，所以 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{E}} \htmlData{tutor-start=20,tutor-end=21}{=} \frac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\mathbf{\htmlData{tutor-start=41,tutor-end=42}{u}}$。在平行四边形中，$\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{B}\htmlData{tutor-start=39,tutor-end=40}{C}} \htmlData{tutor-start=42,tutor-end=43}{=} \mathbf{\htmlData{tutor-start=52,tutor-end=53}{v}}$，因此点 $\htmlData{tutor-start=0,tutor-end=1}{D}$ 相对于 $\htmlData{tutor-start=0,tutor-end=1}{B}$ 的位置向量为 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{B}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{B}\htmlData{tutor-start=39,tutor-end=40}{A}} \htmlData{tutor-start=42,tutor-end=43}{+} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{A}\htmlData{tutor-start=61,tutor-end=62}{D}} \htmlData{tutor-start=64,tutor-end=65}{=} \mathbf{\htmlData{tutor-start=74,tutor-end=75}{u}} \htmlData{tutor-start=77,tutor-end=78}{+} \mathbf{\htmlData{tutor-start=87,tutor-end=88}{v}}$。接下来表示目标向量：对角线向量 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{A}\htmlData{tutor-start=17,tutor-end=18}{C}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{B}\htmlData{tutor-start=39,tutor-end=40}{C}} \htmlData{tutor-start=42,tutor-end=43}{-} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{B}\htmlData{tutor-start=61,tutor-end=62}{A}} \htmlData{tutor-start=64,tutor-end=65}{=} \mathbf{\htmlData{tutor-start=74,tutor-end=75}{v}} \htmlData{tutor-start=77,tutor-end=78}{-} \mathbf{\htmlData{tutor-start=87,tutor-end=88}{u}}$；线段 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}$ 对应的向量 $\overrightarrow{\htmlData{tutor-start=16,tutor-end=17}{E}\htmlData{tutor-start=17,tutor-end=18}{D}} \htmlData{tutor-start=20,tutor-end=21}{=} \overrightarrow{\htmlData{tutor-start=38,tutor-end=39}{B}\htmlData{tutor-start=39,tutor-end=40}{D}} \htmlData{tutor-start=42,tutor-end=43}{-} \overrightarrow{\htmlData{tutor-start=60,tutor-end=61}{B}\htmlData{tutor-start=61,tutor-end=62}{E}} \htmlData{tutor-start=64,tutor-end=65}{=} \htmlData{tutor-start=66,tutor-end=67}{(}\mathbf{\htmlData{tutor-start=75,tutor-end=76}{u}} \htmlData{tutor-start=78,tutor-end=79}{+} \mathbf{\htmlData{tutor-start=88,tutor-end=89}{v}}\htmlData{tutor-start=90,tutor-end=91}{)} \htmlData{tutor-start=92,tutor-end=93}{-} \frac{\htmlData{tutor-start=100,tutor-end=101}{1}}{\htmlData{tutor-start=103,tutor-end=104}{2}}\mathbf{\htmlData{tutor-start=113,tutor-end=114}{u}} \htmlData{tutor-start=116,tutor-end=117}{=} \mathbf{\htmlData{tutor-start=126,tutor-end=127}{v}} \htmlData{tutor-start=129,tutor-end=130}{+} \frac{\htmlData{tutor-start=137,tutor-end=138}{1}}{\htmlData{tutor-start=140,tutor-end=141}{2}}\mathbf{\htmlData{tutor-start=150,tutor-end=151}{u}}$。

记 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{v}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{s}\htmlData{tutor-start=18,tutor-end=19}{,}\mathbf{\htmlData{tutor-start=27,tutor-end=28}{u}}\htmlData{tutor-start=29,tutor-end=34}{\cdot}\mathbf{\htmlData{tutor-start=42,tutor-end=43}{v}}\htmlData{tutor-start=44,tutor-end=45}{=}\htmlData{tutor-start=45,tutor-end=46}{t}$。把 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{9}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{D}\htmlData{tutor-start=6,tutor-end=7}{E}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}$ 分别平方。

详细展开：利用向量模长与数量积的关系 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{a}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \mathbf{\htmlData{tutor-start=27,tutor-end=28}{a}} \htmlData{tutor-start=30,tutor-end=36}{\cdot }\mathbf{\htmlData{tutor-start=44,tutor-end=45}{a}}$，我们将已知长度转化为关于基底向量模长和数量积的方程。设 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{v}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{s}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{v}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{t}$。已知 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=12}{|} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{0}$，故 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{0}$。 对于 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{9}$，有 $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{A}\htmlData{tutor-start=18,tutor-end=19}{C}}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{|}\mathbf{\htmlData{tutor-start=37,tutor-end=38}{v}} \htmlData{tutor-start=40,tutor-end=41}{-} \mathbf{\htmlData{tutor-start=50,tutor-end=51}{u}}\htmlData{tutor-start=52,tutor-end=53}{|}^{\htmlData{tutor-start=55,tutor-end=56}{2}} \htmlData{tutor-start=58,tutor-end=59}{=} \htmlData{tutor-start=60,tutor-end=61}{(}\mathbf{\htmlData{tutor-start=69,tutor-end=70}{v}} \htmlData{tutor-start=72,tutor-end=73}{-} \mathbf{\htmlData{tutor-start=82,tutor-end=83}{u}}\htmlData{tutor-start=84,tutor-end=85}{)} \htmlData{tutor-start=86,tutor-end=92}{\cdot }\htmlData{tutor-start=92,tutor-end=93}{(}\mathbf{\htmlData{tutor-start=101,tutor-end=102}{v}} \htmlData{tutor-start=104,tutor-end=105}{-} \mathbf{\htmlData{tutor-start=114,tutor-end=115}{u}}\htmlData{tutor-start=116,tutor-end=117}{)} \htmlData{tutor-start=118,tutor-end=119}{=} \htmlData{tutor-start=120,tutor-end=121}{|}\mathbf{\htmlData{tutor-start=129,tutor-end=130}{v}}\htmlData{tutor-start=131,tutor-end=132}{|}^{\htmlData{tutor-start=134,tutor-end=135}{2}} \htmlData{tutor-start=137,tutor-end=138}{-} \htmlData{tutor-start=139,tutor-end=140}{2}\mathbf{\htmlData{tutor-start=148,tutor-end=149}{u}} \htmlData{tutor-start=151,tutor-end=157}{\cdot }\mathbf{\htmlData{tutor-start=165,tutor-end=166}{v}} \htmlData{tutor-start=168,tutor-end=169}{+} \htmlData{tutor-start=170,tutor-end=171}{|}\mathbf{\htmlData{tutor-start=179,tutor-end=180}{u}}\htmlData{tutor-start=181,tutor-end=182}{|}^{\htmlData{tutor-start=184,tutor-end=185}{2}} \htmlData{tutor-start=187,tutor-end=188}{=} \htmlData{tutor-start=189,tutor-end=190}{s} \htmlData{tutor-start=191,tutor-end=192}{-} \htmlData{tutor-start=193,tutor-end=194}{2}\htmlData{tutor-start=194,tutor-end=195}{t} \htmlData{tutor-start=196,tutor-end=197}{+} \htmlData{tutor-start=198,tutor-end=199}{1}\htmlData{tutor-start=199,tutor-end=200}{0}\htmlData{tutor-start=200,tutor-end=201}{0}$。因为 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{9}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{t} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{0} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{9}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{8}\htmlData{tutor-start=24,tutor-end=25}{1}$。 对于 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}$，有 $\htmlData{tutor-start=0,tutor-end=1}{|}\overrightarrow{\htmlData{tutor-start=17,tutor-end=18}{E}\htmlData{tutor-start=18,tutor-end=19}{D}}\htmlData{tutor-start=20,tutor-end=21}{|}^{\htmlData{tutor-start=23,tutor-end=24}{2}} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{|}\mathbf{\htmlData{tutor-start=37,tutor-end=38}{v}} \htmlData{tutor-start=40,tutor-end=41}{+} \frac{\htmlData{tutor-start=48,tutor-end=49}{1}}{\htmlData{tutor-start=51,tutor-end=52}{2}}\mathbf{\htmlData{tutor-start=61,tutor-end=62}{u}}\htmlData{tutor-start=63,tutor-end=64}{|}^{\htmlData{tutor-start=66,tutor-end=67}{2}} \htmlData{tutor-start=69,tutor-end=70}{=} \htmlData{tutor-start=71,tutor-end=72}{(}\mathbf{\htmlData{tutor-start=80,tutor-end=81}{v}} \htmlData{tutor-start=83,tutor-end=84}{+} \frac{\htmlData{tutor-start=91,tutor-end=92}{1}}{\htmlData{tutor-start=94,tutor-end=95}{2}}\mathbf{\htmlData{tutor-start=104,tutor-end=105}{u}}\htmlData{tutor-start=106,tutor-end=107}{)} \htmlData{tutor-start=108,tutor-end=114}{\cdot }\htmlData{tutor-start=114,tutor-end=115}{(}\mathbf{\htmlData{tutor-start=123,tutor-end=124}{v}} \htmlData{tutor-start=126,tutor-end=127}{+} \frac{\htmlData{tutor-start=134,tutor-end=135}{1}}{\htmlData{tutor-start=137,tutor-end=138}{2}}\mathbf{\htmlData{tutor-start=147,tutor-end=148}{u}}\htmlData{tutor-start=149,tutor-end=150}{)} \htmlData{tutor-start=151,tutor-end=152}{=} \htmlData{tutor-start=153,tutor-end=154}{|}\mathbf{\htmlData{tutor-start=162,tutor-end=163}{v}}\htmlData{tutor-start=164,tutor-end=165}{|}^{\htmlData{tutor-start=167,tutor-end=168}{2}} \htmlData{tutor-start=170,tutor-end=171}{+} \htmlData{tutor-start=172,tutor-end=173}{2}\htmlData{tutor-start=173,tutor-end=174}{(}\frac{\htmlData{tutor-start=180,tutor-end=181}{1}}{\htmlData{tutor-start=183,tutor-end=184}{2}}\htmlData{tutor-start=185,tutor-end=186}{)}\mathbf{\htmlData{tutor-start=194,tutor-end=195}{u}} \htmlData{tutor-start=197,tutor-end=203}{\cdot }\mathbf{\htmlData{tutor-start=211,tutor-end=212}{v}} \htmlData{tutor-start=214,tutor-end=215}{+} \frac{\htmlData{tutor-start=222,tutor-end=223}{1}}{\htmlData{tutor-start=225,tutor-end=226}{4}}\htmlData{tutor-start=227,tutor-end=228}{|}\mathbf{\htmlData{tutor-start=236,tutor-end=237}{u}}\htmlData{tutor-start=238,tutor-end=239}{|}^{\htmlData{tutor-start=241,tutor-end=242}{2}} \htmlData{tutor-start=244,tutor-end=245}{=} \htmlData{tutor-start=246,tutor-end=247}{s} \htmlData{tutor-start=248,tutor-end=249}{+} \htmlData{tutor-start=250,tutor-end=251}{t} \htmlData{tutor-start=252,tutor-end=253}{+} \frac{\htmlData{tutor-start=260,tutor-end=261}{1}}{\htmlData{tutor-start=263,tutor-end=264}{4}}\htmlData{tutor-start=265,tutor-end=266}{(}\htmlData{tutor-start=266,tutor-end=267}{1}\htmlData{tutor-start=267,tutor-end=268}{0}\htmlData{tutor-start=268,tutor-end=269}{0}\htmlData{tutor-start=269,tutor-end=270}{)} \htmlData{tutor-start=271,tutor-end=272}{=} \htmlData{tutor-start=273,tutor-end=274}{s} \htmlData{tutor-start=275,tutor-end=276}{+} \htmlData{tutor-start=277,tutor-end=278}{t} \htmlData{tutor-start=279,tutor-end=280}{+} \htmlData{tutor-start=281,tutor-end=282}{2}\htmlData{tutor-start=282,tutor-end=283}{5}$。因为 $\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{E}\htmlData{tutor-start=2,tutor-end=3}{=}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{t} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{5} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{2}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{=} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{4}$。 综上得到方程组：$\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{0}\htmlData{tutor-start=6,tutor-end=7}{0} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{t} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{8}\htmlData{tutor-start=16,tutor-end=17}{1}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{5} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{t} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{4}$。

联立两式得 $\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{6}\htmlData{tutor-start=4,tutor-end=5}{,}\htmlData{tutor-start=5,tutor-end=6}{s}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{7}\htmlData{tutor-start=8,tutor-end=9}{3}$。这些量足以确定平行四边形面积。

详细展开：现在我们解由上一步得到的二元一次方程组： (1) $s - 2t + 100 = 81 \implies s - 2t = -19$ (2) $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{t} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{5} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{4} \implies \htmlData{tutor-start=26,tutor-end=27}{s} \htmlData{tutor-start=28,tutor-end=29}{+} \htmlData{tutor-start=30,tutor-end=31}{t} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{9}$ 用方程 (2) 减去方程 (1)：$\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{s} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{t}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{s} \htmlData{tutor-start=13,tutor-end=14}{-} \htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{t}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{9} \htmlData{tutor-start=25,tutor-end=26}{-} \htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{9}\htmlData{tutor-start=31,tutor-end=32}{)}$，即 $\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{t} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{1}\htmlData{tutor-start=6,tutor-end=7}{3}\htmlData{tutor-start=7,tutor-end=8}{8}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{6}$。 将 $\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{6}$ 代入方程 (2)：$\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{6} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{9}$，解得 $\htmlData{tutor-start=0,tutor-end=1}{s} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{7}\htmlData{tutor-start=5,tutor-end=6}{3}$。 因此，我们求得 $\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{v}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{4}\htmlData{tutor-start=31,tutor-end=32}{6}$，以及 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{v}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{7}\htmlData{tutor-start=20,tutor-end=21}{3}$（即 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{v}}\htmlData{tutor-start=11,tutor-end=12}{|} \htmlData{tutor-start=13,tutor-end=14}{=} \sqrt{\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=23}{3}}$）。这些数值确定了平行四边形邻边的夹角余弦值和另一边长，从而唯一确定了形状和面积。

平行四边形面积等于 $\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=17}{\times}\mathbf{\htmlData{tutor-start=25,tutor-end=26}{v}}\htmlData{tutor-start=27,tutor-end=28}{|}$。

详细展开：平行四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 的面积 $\htmlData{tutor-start=0,tutor-end=1}{S}$ 等于邻边向量叉积的模，即 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{|}\mathbf{\htmlData{tutor-start=13,tutor-end=14}{u}} \htmlData{tutor-start=16,tutor-end=23}{\times }\mathbf{\htmlData{tutor-start=31,tutor-end=32}{v}}\htmlData{tutor-start=33,tutor-end=34}{|}$。在二维或三维空间中，利用拉格朗日恒等式（或向量面积公式），有 $\htmlData{tutor-start=0,tutor-end=1}{S}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{|}\mathbf{\htmlData{tutor-start=17,tutor-end=18}{u}}\htmlData{tutor-start=19,tutor-end=20}{|}^{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=26}{|}\mathbf{\htmlData{tutor-start=34,tutor-end=35}{v}}\htmlData{tutor-start=36,tutor-end=37}{|}^{\htmlData{tutor-start=39,tutor-end=40}{2}} \htmlData{tutor-start=42,tutor-end=43}{-} \htmlData{tutor-start=44,tutor-end=45}{(}\mathbf{\htmlData{tutor-start=53,tutor-end=54}{u}} \htmlData{tutor-start=56,tutor-end=62}{\cdot }\mathbf{\htmlData{tutor-start=70,tutor-end=71}{v}}\htmlData{tutor-start=72,tutor-end=73}{)}^{\htmlData{tutor-start=75,tutor-end=76}{2}}$。 代入已知数值：$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{u}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{0}$，$\htmlData{tutor-start=0,tutor-end=1}{|}\mathbf{\htmlData{tutor-start=9,tutor-end=10}{v}}\htmlData{tutor-start=11,tutor-end=12}{|}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \htmlData{tutor-start=19,tutor-end=20}{s} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{7}\htmlData{tutor-start=24,tutor-end=25}{3}$，$\mathbf{\htmlData{tutor-start=8,tutor-end=9}{u}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\mathbf{\htmlData{tutor-start=25,tutor-end=26}{v}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{t} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{4}\htmlData{tutor-start=35,tutor-end=36}{6}$。 计算：$\htmlData{tutor-start=0,tutor-end=1}{S}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0} \htmlData{tutor-start=12,tutor-end=19}{\times }\htmlData{tutor-start=19,tutor-end=20}{7}\htmlData{tutor-start=20,tutor-end=21}{3} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{6}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{7}\htmlData{tutor-start=34,tutor-end=35}{3}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{0} \htmlData{tutor-start=38,tutor-end=39}{-} \htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{6} \htmlData{tutor-start=45,tutor-end=46}{=} \htmlData{tutor-start=47,tutor-end=48}{5}\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{8}\htmlData{tutor-start=50,tutor-end=51}{4}$。 开平方得 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \sqrt{\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{8}\htmlData{tutor-start=13,tutor-end=14}{4}}$。因为 $\htmlData{tutor-start=0,tutor-end=1}{7}\htmlData{tutor-start=1,tutor-end=2}{0}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{=}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{9}\htmlData{tutor-start=9,tutor-end=10}{0}\htmlData{tutor-start=10,tutor-end=11}{0}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{7}\htmlData{tutor-start=14,tutor-end=15}{2}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{7}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{+}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{4}\htmlData{tutor-start=32,tutor-end=33}{9}\htmlData{tutor-start=33,tutor-end=34}{0}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{2}\htmlData{tutor-start=37,tutor-end=38}{8}\htmlData{tutor-start=38,tutor-end=39}{0}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{4}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{5}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{8}\htmlData{tutor-start=45,tutor-end=46}{4}$，所以 $\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{7}\htmlData{tutor-start=5,tutor-end=6}{2}$。 故平行四边形 $\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C}\htmlData{tutor-start=3,tutor-end=4}{D}$ 的面积为 72。

$\htmlData{tutor-start=0,tutor-end=1}{5}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{2}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{3}^{\htmlData{tutor-start=17,tutor-end=18}{2}}$，所以三角形为直角三角形，斜边长 13。取直角顶点为原点，另两点可设为 $\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{0}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{)}$。

详细展开：首先观察给定的三边长 $\htmlData{tutor-start=0,tutor-end=1}{5}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{3}$。我们需要判断三角形的形状，因为特殊三角形（如直角三角形）的中心位置有简便性质。计算两较短边的平方和：$\htmlData{tutor-start=0,tutor-end=1}{5}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{5} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{4}\htmlData{tutor-start=24,tutor-end=25}{4} \htmlData{tutor-start=26,tutor-end=27}{=} \htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{6}\htmlData{tutor-start=30,tutor-end=31}{9}$。再计算最长边的平方：$\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{3}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{6}\htmlData{tutor-start=11,tutor-end=12}{9}$。因为 $\htmlData{tutor-start=0,tutor-end=1}{5}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{3}^{\htmlData{tutor-start=21,tutor-end=22}{2}}$，根据勾股定理的逆定理，该三角形是以长度为 $\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{3}$ 的边为斜边的直角三角形。为了后续计算方便，我们建立平面直角坐标系：将直角顶点设为原点 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$，两条直角边分别落在 $\htmlData{tutor-start=0,tutor-end=1}{x}$ 轴和 $\htmlData{tutor-start=0,tutor-end=1}{y}$ 轴的正半轴上。因此，另外两个顶点的坐标可以设为 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}$。这样，三个顶点的坐标就完全确定了。

直角三角形的垂心就是直角顶点，外心是斜边中点，故二者距离等于斜边的一半。

详细展开：在直角三角形中，垂心 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 的位置非常特殊，它恰好位于直角顶点处。在我们建立的坐标系中，直角顶点即为原点，所以垂心坐标为 $\htmlData{tutor-start=0,tutor-end=1}{H}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$。接下来确定外心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 的位置。根据几何性质，直角三角形的外接圆圆心（外心）位于斜边的中点。斜边连接点 $\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{5}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}$ 和 $\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{)}$。利用中点坐标公式，外心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 的横坐标为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{5}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{0}}{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{=} \frac{\htmlData{tutor-start=22,tutor-end=23}{5}}{\htmlData{tutor-start=25,tutor-end=26}{2}}$，纵坐标为 $\frac{\htmlData{tutor-start=6,tutor-end=7}{0}\htmlData{tutor-start=7,tutor-end=8}{+}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{2}}{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \htmlData{tutor-start=17,tutor-end=18}{6}$。即 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{(}\frac{\htmlData{tutor-start=8,tutor-end=9}{5}}{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{,} \htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{)}$。现在计算垂心 $\htmlData{tutor-start=0,tutor-end=1}{H}$ 到外心 $\htmlData{tutor-start=0,tutor-end=1}{O}$ 的距离 $\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{H}$。根据两点间距离公式：$\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{H} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{3}}{\htmlData{tutor-start=15,tutor-end=16}{2}}$。这里也可以直接利用结论：直角三角形垂心到外心的距离等于斜边长度的一半，即 $\frac{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{3}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{=} \htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{5}$。