A,B,C

$\because AB=AC$，$\angle BAC=90^{\circ}$，
$\therefore \angle B=\angle C=45^{\circ}$，
$\because$直角$\angle EPF$的顶点$P$是$BC$的中点，
$\therefore AP\bot BC$，$AP=CP=BP=\frac{1}{2}BC$，$\angle EAP=\angle CAP=\frac{1}{2}\angle BAC=45^{\circ}$，
$\therefore \angle EAP=\angle C$，$\angle EPF=\angle APC=90^{\circ}$，
$\therefore \angle APE=\angle CPF=90^{\circ}-\angle APF$，
在$\triangle APE$和$\triangle CPF$中，
$\left\{\begin{array}{l}{∠EAP=∠C}\\{AP=CP}\\{∠APE=∠CPF}\end{array}\right.$，
$\therefore \triangle APE$≌$\triangle CPF\left(ASA\right)$，
$\therefore AE=CF$，$EP=FP$，
$\therefore \triangle EPF$是等腰直角三角形，
故$A$正确，$B$正确；
$\because S_{\triangle APE}=S_{\triangle CPF}$，且$S_{\triangle APC}=S_{\triangle APB}=\frac{1}{2}S_{\triangle ABC}$，
$\therefore S_{四边形AEPF}=S_{\triangle APE}+S_{\triangle APF}=S_{\triangle CPE}+S_{\triangle APF}=S_{\triangle APC}=\frac{1}{2}S_{\triangle ABC}$，
故$C$正确；
$\because AE=CF$，
$\therefore BE+CF=BE+AE=AB$，
$\because$点$E$不与$A$，$B$重合
$\because BP\neq EP$，
$\therefore \sqrt{2}BP\neq \sqrt{2}EP$，
$\because AB=\sqrt{A{P}^{2}+B{P}^{2}}=\sqrt{2}BP$，$EF=\sqrt{E{P}^{2}+F{P}^{2}}=\sqrt{2}EP$，
$\therefore AB\neq EF$，
$\therefore BE+CF\neq EF$，
故$D$错误，
故选：$ABC$.