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HuangHai
2025-08-30 18:35:01 +08:00
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- $AE=CF,\angle ACF=\angle EAB,AB=AC$
$\therefore \triangle CAF \cong \triangle AEB$
$\therefore \angle ABE=\angle CAF$
开始导角:
$\angle 1=\angle 2,\angle 1+\angle 3=60^{\circ}$
$\therefore \angle 2+\angle 3=60^{\circ}$
$\therefore \angle APB=120^{\circ}$
同时知道,此时$AB=6$
$\therefore$符合定角定弦规则,画出外接圆,设圆心为$O$
连接 $OA,OB,OC$
因为$\angle APB$一直等于$120^{\circ}$,所以$P$的轨迹一定是按圆的轨迹来运动,当$P$运动到$O,P,C$三点共线时,$CP$最短。
同时可以证明$\triangle AOC$是直角三角形,且$\angle OCA=30^{\circ}$
可求$CP=3\sqrt{3}$