如图,AB∥CD,点E在BC上,且CD=CE,∠D=74°,则∠B的度数为( ) A.

如图,ABCD,点EBC上,且CD=CED=74°,则B的度数为(  )

A68°B32°C22°D16°

答案

B【考点】平行线的性质;等腰三角形的性质.

【分析】根据等腰三角形两底角相等求出C的度数,再根据两直线平行,内错角相等解答即可.

【解答】解:CD=CE

∴∠D=DEC

∵∠D=74°

∴∠C=180°74°×2=32°

ABCD

∴∠B=C=32°

故选B

 

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