人教版初三数学压轴题解题模型之角平分线模型(最新整理)

E F

A

D

E

F

B C

【例4】已知如图在△ABC 中，∠ACB=90°，CD⊥AB 于D，∠A 的平分线交CD 于F，BC 于E，过点E 作EH⊥AB 于H．

求证：（1）CF=EH．（2）四边形CEHF 是菱形．

【参考答案】（1）提示：先证△AEC≌△ACF，得到CE=EH，∠CEA=∠HEA

再借助平行线得到∠CFE=∠HEA，∠CFE=∠CEA，得到CE=CF，进而得到CF=EH

（2）联结FH，由CF 平行且等于EH，得到CEHF 是平行四边形

再由EC=EH 得到CEHF 是菱形

课堂练习此环节设计时间在30 分钟左右（20 分钟练习＋10 分钟互动讲解）。

1．已知：如图，平行四边形ABCD 各角的平分线分别相交于点E，F，G，H，

求证：四边形EFGH 是矩形．

1

1 E F

1

2

【参考答案】证明：∵四边形 ABCD 是平行四边形，所以 AD ∥BC

所以∠DAB +∠ABC =180°

因为∠HAB = ∠BAD , ∠HBA = ∠ABC ，

2 2

1

所以∠HAB + ∠HBA = 所以∠H =90°

(∠DAB + ∠ABC ) = 90︒

2

同理：∠F =∠HEF =∠HGF =90°

所以四边形 EFGH 是矩形

2. 已知：如图， ∠BAD = ∠CAD , AB > AC , CD ⊥ AD 于点 D , H 是 BC 中点，

1

求证： DH =

(AB - AC )．

2

A

B

【参考答案】提示：延长 CD 交 AB 于点 E ，用全等和中位线性质去证明

3. 如图，已知∠BAC =90°，AD ⊥BC 于点 D ，∠1=∠2，EF ∥BC 交 AC 于点 F ．试说明 AE =CF ．

A

B

D

C

D C

H

C

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