已知：如圖，$\angle C=\angle C' = 90^{\circ}，$$AC = A'C'，$$AD$平分$\angle BAC，$$A'D'$平分$\angle B'A'C'，$$AD = A'D'。$
求證：$\triangle ABC\cong\triangle A'B'C'。$
證明：在$Rt\triangle ACD$和$Rt\triangle A'C'D'$中，$\begin{cases}AC = A'C'\\AD = A'D'\end{cases}，$$\therefore Rt\triangle ACD\cong Rt\triangle A'C'D'(HL)，$$\therefore\angle CAD=\angle C'A'D'。$
$\because AD$平分$\angle BAC，$$A'D'$平分$\angle B'A'C'，$$\therefore\angle CAB = 2\angle CAD，$$\angle C'A'B' = 2\angle C'A'D'，$$\therefore\angle CAB=\angle C'A'B'。$
在$\triangle ABC$與$\triangle A'B'C'$中，$\begin{cases}\angle CAB=\angle C'A'B'\\AC = A'C'\\\angle C=\angle C'\end{cases}，$$\therefore\triangle ABC\cong\triangle A'B'C'(ASA)。$
