题目内容

如图,在△ABC中,AB=ACDBC的中点,四边形ABDE是平行四边形.

(1)求证:四边形ADCE是矩形;

(2)若ACDE交于点O,四边形ADCE的面积为CD=4,求∠AOD的度数.

 (1)∵在△ABC中,AB=ACDBC的中点,

ADBC

∴∠AOD =90°······························ 1分

∵四边形ABDE是平行四边形

AE=BDAE∥BD····························· 2分

DBC的中点,

DC=BD······························· 3分

AE=DCAE∥DC

∴四边形ADCE是平行四边形······················ 4分

∴四边形ADCE是矩形························· 5分

 (2)∵矩形ADCE的面积=CD=4

AD=······························ 6分

∴在RtABC中,tanCAD=   ··············· 7分 

∴∠CAD=30°

又∵四边形ADCE是矩形

OD=OA

∴∠AOD=120°···························· 8分

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