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1. 如图,在△ABC 中,∠A = 100°,∠ABC = 40°,BD 是△ABC 的角平分线,延长 BD 至点 E,使 DE = AD,连接 EC. 求证:BC = AB + CE.
答案
证明:在 BC 上截取 $ BM = BA $,连接 DM.
$ \because BD $ 平分 $ \angle ABC $,
$ \therefore \angle ABD = \angle MBD = \frac{1}{2} \angle ABC = 20^{\circ} $.
$ \because BD = BD $,
$ \therefore \triangle ABD \cong \triangle MBD $,
$ \therefore DA = DM $, $ \angle ADB = \angle BDM = 180^{\circ} - \angle A - \angle ABD = 60^{\circ} $,
$ \therefore \angle EDC = \angle ADB = 60^{\circ} $,
$ \angle MDC = 60^{\circ} $,
$ \therefore \angle EDC = \angle MDC $.
$ \because DE = AD $,
$ \therefore DE = DM $,
$ \therefore \triangle EDC \cong \triangle MDC $,
$ \therefore MC = EC $,
$ \therefore BC = BM + MC = AB + EC $.
$ \because BD $ 平分 $ \angle ABC $,
$ \therefore \angle ABD = \angle MBD = \frac{1}{2} \angle ABC = 20^{\circ} $.
$ \because BD = BD $,
$ \therefore \triangle ABD \cong \triangle MBD $,
$ \therefore DA = DM $, $ \angle ADB = \angle BDM = 180^{\circ} - \angle A - \angle ABD = 60^{\circ} $,
$ \therefore \angle EDC = \angle ADB = 60^{\circ} $,
$ \angle MDC = 60^{\circ} $,
$ \therefore \angle EDC = \angle MDC $.
$ \because DE = AD $,
$ \therefore DE = DM $,
$ \therefore \triangle EDC \cong \triangle MDC $,
$ \therefore MC = EC $,
$ \therefore BC = BM + MC = AB + EC $.
2. 如图,在△ABC 中,∠ABC = 60°,△ABC 的两条角平分线 AD,CE 相交于点 P.
(1)求∠CPD 的度数;
(2)求证:AC = AE + CD.
(1)求∠CPD 的度数;
(2)求证:AC = AE + CD.
答案
解:(1)$ \because AD $, $ CE $ 分别平分 $ \angle BAC $, $ \angle ACB $,
$ \therefore \angle PAC = \frac{1}{2} \angle BAC $,
$ \angle PCA = \frac{1}{2} \angle ACB $,
$ \therefore \angle APC = 180^{\circ} - (\angle PAC + \angle PCA) $
$ = 180^{\circ} - \frac{1}{2} (\angle BAC + \angle ACB) $
$ = 120^{\circ} $,
$ \therefore \angle CPD = 60^{\circ} $;
(2)在 AC 上截取 $ CF = CD $,连接 PF.
$ \because \angle PCD = \angle PCF $, $ PC = PC $,
$ \therefore \triangle PCD \cong \triangle PCF $,
$ \therefore \angle CPF = \angle CPD = 60^{\circ} $,
$ \therefore \angle APF = 120^{\circ} - 60^{\circ} = 60^{\circ} $.
又 $ \angle APE = \angle CPD = 60^{\circ} $,
$ \therefore \angle APE = \angle APF $.
$ \because \angle BAD = \angle CAD $, $ AP = AP $,
$ \therefore \triangle APE \cong \triangle APF $,
$ \therefore AF = AE $.
$ \because CD = CF $,
$ \therefore AC = AF + CF = AE + CD $.
$ \therefore \angle PAC = \frac{1}{2} \angle BAC $,
$ \angle PCA = \frac{1}{2} \angle ACB $,
$ \therefore \angle APC = 180^{\circ} - (\angle PAC + \angle PCA) $
$ = 180^{\circ} - \frac{1}{2} (\angle BAC + \angle ACB) $
$ = 120^{\circ} $,
$ \therefore \angle CPD = 60^{\circ} $;
(2)在 AC 上截取 $ CF = CD $,连接 PF.
$ \because \angle PCD = \angle PCF $, $ PC = PC $,
$ \therefore \triangle PCD \cong \triangle PCF $,
$ \therefore \angle CPF = \angle CPD = 60^{\circ} $,
$ \therefore \angle APF = 120^{\circ} - 60^{\circ} = 60^{\circ} $.
又 $ \angle APE = \angle CPD = 60^{\circ} $,
$ \therefore \angle APE = \angle APF $.
$ \because \angle BAD = \angle CAD $, $ AP = AP $,
$ \therefore \triangle APE \cong \triangle APF $,
$ \therefore AF = AE $.
$ \because CD = CF $,
$ \therefore AC = AF + CF = AE + CD $.
3. (教材变式)如图,AD//BC,E 是 CD 上一点,且∠1 = ∠2,∠3 = ∠4. 求证:AB = AD + BC.
答案
证明:方法一:(补短法)延长 AD 至点 F,使 $ AF = AB $,连接 EF,
先证 $ \triangle AEF \cong \triangle AEB $,再证 $ \triangle EDF \cong \triangle ECB $, $ DF = BC $ 即可.
方法二:(截长法)在 AB 上截取 $ AF = AD $,连接 EF.
先证 $ \triangle AED \cong \triangle AEF $,再证 $ \triangle BEC \cong \triangle BEF $ 即可.
方法三:(作垂线)过点 E 分别向 AD, AB, BC 作垂线,垂足分别为点 G, M, N, $ EG = EM = EN $.
再证 $ AG = AM $, $ BM = BN $,
$ \triangle DEG \cong \triangle CEN $, $ DG = CN $,
$ \therefore AD + BC = (AG - DG) + (BN + CN) = AG + BN = AM + BM = AB $.
先证 $ \triangle AEF \cong \triangle AEB $,再证 $ \triangle EDF \cong \triangle ECB $, $ DF = BC $ 即可.
方法二:(截长法)在 AB 上截取 $ AF = AD $,连接 EF.
先证 $ \triangle AED \cong \triangle AEF $,再证 $ \triangle BEC \cong \triangle BEF $ 即可.
方法三:(作垂线)过点 E 分别向 AD, AB, BC 作垂线,垂足分别为点 G, M, N, $ EG = EM = EN $.
再证 $ AG = AM $, $ BM = BN $,
$ \triangle DEG \cong \triangle CEN $, $ DG = CN $,
$ \therefore AD + BC = (AG - DG) + (BN + CN) = AG + BN = AM + BM = AB $.
