$證明:連接AD$
$在\triangle ABD和\triangle ACD中，$
$\because \left\{\begin{array}{l}AB=AC\\ DB=DC\\ AD=AD\end{array}\right.， $
$\therefore \triangle ABD≌ \triangle ACD\left(SSS\right)， $
$\therefore \angle BAD=\angle CAD， $
$\therefore AD平分\angle BAC， $
$\because DE\bot AB，DF\bot AC， $
$\therefore DE=DF $
