如图，在正方形$ABCD$的边$BC$的延长线上取一点$E$，使$CE=AC$，连接$AE$交$CD$于$F$，则$\angle AFC$等于（ ）A. $112.5^{\circ}$B. $120^{\circ}$C. $135^{\circ}$D. $145^{\circ}$

$\because $四边形$ABCD$为正方形，

$\therefore \angle ACD=90^{\circ}$，

$\therefore \angle DCE=90^{\circ}$，

又$\because AC$是正方形$ABCD$的对角线，

$\therefore \angle ACF=45^{\circ}$，

$\therefore \angle ACE=\angle DCE+\angle ACF=135^{\circ}$，

$\because CE=CA$，

$\therefore \angle FAC=\angle E=\frac{1}{2}(180^{\circ}-135^{\circ})=22.5^{\circ}$

$\therefore \angle AFD=\angle FAC+\angle ACF=22.5^{\circ}+45^{\circ}=67.5^{\circ}$，

$\therefore \angle AFC=180^{\circ}-67.5^{\circ}=112.5^{\circ}$，

故选：$A$.

标签：circ,ABCD,BC