1. (教材变式)如图,在$△ABC$中,AD 平分$∠BAC$交 BC 于点 D,$DE⊥AB$,垂足为 E,$S_{△ABC}= 9,$$DE= 2,AB= 5$.求 AC 的长.

解:过点 D 作 $ DF \perp AC $,垂足为 F.
$ \because AD $ 平分 $ \angle BAC $, $ DE \perp AB $,
$ \therefore DF = DE = 2 $,
$ \therefore S_{\triangle ABC} = S_{\triangle ABD} + S_{\triangle ADC} $
$ = \frac{1}{2}AB \cdot DE + \frac{1}{2}AC \cdot DF $
$ = 5 + AC = 9 $,
$ \therefore AC = 4 $.

2. 如图,$PA= PB,∠1+∠2= 180^{\circ }$. 求证:OP 平分$∠AOB.$

证明:过点 P 分别作 $ PE \perp AO $, $ PF \perp OB $,垂足分别为 E,F,
$ \therefore \angle PEA = \angle PFB = 90^{\circ} $.
$ \because \angle 1 + \angle 2 = 180^{\circ} $,
$ \angle 2 + \angle PBO = 180^{\circ} $,
$ \therefore \angle 1 = \angle PBO $.
又 $ \because PA = PB $,
$ \therefore \triangle PAE \cong \triangle PBF (AAS) $,
$ \therefore PE = PF $,
$ \therefore OP $ 平分 $ \angle AOB $.

3. 如图,$∠ACB= ∠DCE= 90^{\circ },BC= AC,EC= DC$,直线 AD 与 BE 交于点 F,连接 CF.
(1)求证:$BF⊥AD;$
(2)求$∠AFC$的度数.

解:(1) $ \because \angle ACB = \angle DCE = 90^{\circ} $,
$ \therefore \angle BCE = \angle ACD $.
$ \because BC = AC $, $ EC = DC $,
$ \therefore \triangle BCE \cong \triangle ACD $,
$ \therefore \angle A = \angle B $,
$ \therefore \angle AFB = \angle ACB = 90^{\circ} $,
$ \therefore BF \perp AD $;
(2)过点 C 作直线 AD,BE 的垂线,垂足分别为 M,N.
由(1)知 $ \triangle BCE \cong \triangle ACD $,
$ \therefore \angle B = \angle A $.
$ \because \angle CNB = \angle M = 90^{\circ} $, $ BC = AC $,
$ \therefore \triangle BCN \cong \triangle ACM $,
$ \therefore CM = CN $.
$ \because CM \perp AD $, $ CN \perp BE $,
$ \therefore FC $ 平分 $ \angle BFD $.
$ \because \angle BFD = \angle AFB = 90^{\circ} $,
$ \therefore \angle BFC = \frac{1}{2} \angle BFD = 45^{\circ} $,
$ \therefore \angle AFC = 90^{\circ} + 45^{\circ} = 135^{\circ} $.