《2025年伴你学山西九年级数学上册北师大版》

14. (本题10分)如图,在正方形$ABCD$中,点$P是边BC$上一点,$BE\perp AP$,$DF\perp AP$,垂足分别是点$E$,$F$.
(1)求证：$EF = AE - BE$；
(2)连接$BF$,如果$\frac{AF}{BF} = \frac{DF}{AD}$,求证：$EF = EP$.

答案：
证明:
(1)如图,
∵四边形 ABCD 为正方形,
$\therefore AB=AD,\angle BAD=90^{\circ}$.
$\because BE\perp AP,DF\perp AP$,
$\therefore \angle BEA=\angle AFD=90^{\circ}$.
$\because \angle1+\angle2=90^{\circ},\angle2+\angle3=90^{\circ}$,
$\therefore \angle1=\angle3$.

在$\triangle ABE$和$\triangle DAF$中,
$\begin{cases}\angle BEA=\angle AFD,\\\angle1=\angle3,\\AB=DA,\end{cases}$
$\therefore \triangle ABE\cong\triangle DAF(AAS),\therefore BE=AF$,
$\therefore EF=AE-AF=AE-BE$.
(2)如图,连接 BF.$\because \frac{AF}{BF}=\frac{DF}{AD},AF=BE$,
$\therefore \frac{BE}{BF}=\frac{DF}{AD},\therefore \frac{BE}{DF}=\frac{BF}{AD}$,
$\therefore Rt\triangle BEF\backsimRt\triangle DFA$,
$\therefore \angle4=\angle3.\because \angle1=\angle3,\therefore \angle4=\angle1$.
$\because \angle5=\angle1,\therefore \angle4=\angle5$,
即 BE 平分$\angle FBP$.
$\because BE\perp EP,\therefore EF=EP$.