$证明:连接AD$
$在\triangle ABD和\triangle ACD中,$
$\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 $