如图,在△ABC中,AB=AC,D为边BC的中点,四边形ABDE是平行四边形,AC

如图,在△ABC中,AB=ACD为边BC的中点,四边形ABDE是平行四边形,ACDE相交于点O

小题1.求证:四边形ADCE是矩形.

小题2.若∠AOE=60°,AE=2,求矩形ADCE对角线的长.

答案

1.证明:∵四边形ABDE是平行四边形,

AB=DE

又∵AB=AC

DE=AC

AB=ACDBC中点,

∴∠ADC=90°,

又∵DBC中点,

CD=BD

CDAECD=AE

∴四边形AECD是平行四边形,

又∴∠ADC=90°,

∴四边形ADCE是矩形.

2.解:∵四边形ADCE是矩形,

AO=EO,∵∠AOE=60°

∴△AOE为等边三角形,

AO=AE=2

AC=2OA=4

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