题目内容

【题目】在四边形ABCD中,AB=CDEF分别为边BCAD的中点,AECD,延长BACD,分别与EF的延长线交于点GH,连接AHED.

(1)求证:AHED

(2)求证:AE=AG.

【答案】(1)(2)见解析.

【解析】

(1)证AEF≌△DHF. AE=DH.AEDH得四边形AEDH是平行四边形.(2)连接AC,设AC的中点为O,连接OEOF.根据三角形中位线性质得OFCDOF=CDOEABOE=ABOE=OF.OFE=∠OEF,∠DHE =∠AGE. AEDHDHE =∠AEG. 所以AGE =∠AEG.

(1)∵AECD

∴∠AEF=∠DHF,∠FAE=∠FDH.

AF=FD

∴△AEF≌△DHF.

AE=DH.

AEDH

∴四边形AEDH是平行四边形.

AHED.

(2)连接AC,设AC的中点为O,连接OEOF.

EF分别为边BCAD的中点,

OFCDOF=CDOEABOE=AB.

∴∠OFE=∠DHE,∠OEF=∠AGE.

AB=CD

OE=OF.

∴∠OFE=∠OEF.

∴∠DHE =∠AGE.

AEDH

∴∠DHE =∠AEG.

∴∠AGE =∠AEG.

AE=AG.

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