题目内容
如图,在平行四边形ABCD中,M是CD的中点,AB=2BC,BM=
,AM=
,则CD的长为( )




A.![]() | B.![]() | C.![]() | D.![]() |
D.
试题分析:因为M为CD中点,
∴CM=DM=


∴∠DAM=∠DMA,∠CBM=∠CMB,
∵∠C+∠D=180°,
∴∠C=2∠DMA,∠D=2∠CMB
∴∠DMA+∠CMB=

∴∠AMB=180°-(∠DMA+∠CMB)=90°,
即△MAB为直角三角形,
∵BM=


∴CD=AB=

故选D.

练习册系列答案
相关题目