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[几何] 证明广义西摩松线的一个线段长度比例关系

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TSC999 Posted 2023-3-7 20:06 |Read mode
P 是三角形 ABC 外接圆上的一点。从 P 点向三角形各边引倾角相等的斜线 PD、PE、PF,则 D、E、F 三点共线,这条线叫广义西摩松线。现在要求证明:\( \frac{ED}{EF}=\frac{BA}{BC}\frac{PC}{PA}\)

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Ly-lie Posted 2023-3-7 20:39
$P$即为密克点,有$ \triangle APC\sim \triangle FPD,\frac{PC}{PA}=\frac{PD}{PF} $,且\[ \frac{PD}{ED}\cdot  \frac{EF}{PF}=\frac{\sin \angle PED}{\sin \angle DPE}\cdot \frac{\sin \angle FPE}{\sin \angle FEP}=\frac{\sin \angle BCP}{\sin \angle ACB}\cdot \frac{\sin \angle BAC}{\sin \angle BCP}=\frac{BC}{BA} \]

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