如图, $\mathrm{AB}=\mathrm{CD}, \mathrm{DE} \perp \mathrm{AC}, \mathrm{BF} \perp \mathrm{AC}, \mathrm{E}, \mathrm{F}$ 是垂足, $\mathrm{AE}=\mathrm{CF}$, 求证: $\mathrm{AB} / / \mathrm{CD}$.

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证明: $\because A E=C F$,
$\therefore A E+E F=C F+E F$, 即 $A F=E C$.

又 $\because B F \perp A C, D E \perp A C$,

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

在 Rt $\triangle A B F$ 与 Rt $\triangle C D E$ 中,

$\because\left\{\begin{array}{l}A F=C E \\ A B=C D\end{array}\right.$,

$\therefore \mathrm{Rt} \triangle A B F \xlongequal{ } \mathrm{Rt} \triangle C D E(\mathrm{HL})$,

$\therefore \angle C=\angle A$,

$\therefore A B / / C D$