1.(教材变式)如图,$AD⊥BC$于点 D,$CE⊥AB$于点 E,AD,CE 交于点 F.
(1)求证:$∠A= ∠C;$
(2)求证:$∠AFE= ∠B$.

证明: (1) $\because AD \perp BC, CE \perp AB$,
$\therefore \angle ADB = \angle CEB = 90^{\circ}$,
$\therefore \angle A + \angle B = \angle C + \angle B = 90^{\circ}$,
$\therefore \angle A = \angle C$;
(2) $\because \angle AEF = \angle ADB = 90^{\circ}$,
$\therefore \angle AFE + \angle A = \angle B + \angle A = 90^{\circ}$,
$\therefore \angle AFE = \angle B$.

2.如图,在四边形 ABCD 中,$∠A= ∠C= 90^{\circ }$,BE 平分$∠ABC$交 AD 于点 E,DF 平分$∠ADC$交 BC 于点 F,BE,CD 的延长线交于点 G.
(1)求证:$∠ABC+∠ADC= 180^{\circ };$
(2)求证:$BG// DF$.

证明: (1) 连接 $BD$.
可得 $\angle ABC + \angle ADC + \angle A + \angle C = 180^{\circ} \times 2 = 360^{\circ}$.
$\because \angle A = \angle C = 90^{\circ}$,
$\therefore \angle ABC + \angle ADC = 180^{\circ}$;
(2) $\because BE$ 平分 $\angle ABC$, $DF$ 平分 $\angle ADC$,
$\therefore \angle CBG = \frac{1}{2} \angle ABC$,
$\angle CDF = \frac{1}{2} \angle ADC$,
$\therefore \angle CBG + \angle CDF = \frac{1}{2} (\angle ABC + \angle ADC) = 90^{\circ}$.
$\because \angle CFD + \angle CDF = 90^{\circ}$,
$\therefore \angle CBG = \angle CFD$,
$\therefore DF // BG$.

3.如图,点 B,D,C 在同一直线上,$∠ADE= ∠B= ∠C$.
(1)求证:$∠CDE= ∠A;$
(2)求证:$∠ADB= ∠E$.

证明: (1) $\because \angle ADC = \angle ADE + \angle CDE = \angle A + \angle B$,
又 $\because \angle ADE = \angle B$,
$\therefore \angle CDE = \angle A$;
(2) $\because \angle BDE = \angle ADB + \angle ADE = \angle E + \angle C$,
$\angle ADE = \angle C$,
$\therefore \angle ADB = \angle E$.