$證明:\because \angle ADB+\angle ADE+\angle EDC={180}^{\circ }$
$\angle B+\angle BAD+\angle ADB={180}^{\circ }$
$又\because \angle BAD=\angle EDC$
$\therefore \angle B=\angle ADE$
$\because \angle ADE=\angle ADB $
$\therefore \angle B=\angle ADB $
$\therefore AB=AD $
$\because \angle BAD=\angle EAC $
$\therefore \angle BAD+\angle DAC=\angle EAC+\angle DAC $
$\therefore \angle BAC=\angle DAE $
$在\triangle ABC和\triangle ADE中 $
$\left \{ {{\begin{array}{ll} {\angle B=\angle ADE} \\\ {AB=AD} \\\ {\angle BAC=\angle DAE} \end{array}}} \right . $
$\therefore \triangle ABC\cong \triangle ADE $
