如图, 已知 $A E \perp A B, A F \perp A C, A E=A B, A F=A C, B F$ 与 $C E$ 相交于点 $M$.

<ImageHere>

(1) 求证: $E C=B F$;

(2) 求证: $E C \perp B F$.
(1) 证明: $\because A E \perp A B, \quad A F \perp A C$,

$\therefore \angle B A E=\angle C A F=90^{\circ}$,

$\therefore \angle B A E+\angle B A C=\angle C A F+\angle B A C$, 即 $\angle E A C=\angle B A F$,

在 $\triangle A B F$ 和 $\triangle A E C$ 中,

$\left\{\begin{aligned} A B & =A E \\ \angle E A C & =\angle B A F \\ A C & =A F\end{aligned}\right.$

$\therefore \triangle A B F \cong \triangle A E C(\mathrm{SAS})$,
$\therefore E C=B F ;$

(2) 证明：设 $A B$ 与 $C E$ 交于点 $D$,

$\because \triangle A B F \cong \triangle A E C$

$\therefore \angle A E C=\angle A B F$

$\because A E \perp A B$,

$\therefore \angle B A E=90^{\circ}$,

$\therefore \angle A E C+\angle A D E=90^{\circ}$,

$\because \angle A D E=\angle B D M （$ 对顶角相等 ），

$\therefore \angle A B F+\angle B D M=90^{\circ}$,

在 $\triangle B D M$ 中, $\angle B M D=180^{\circ}-\angle A B F-\angle B D M=180^{\circ}-90^{\circ}=90^{\circ}$,

$\therefore E C \perp B F$.