如图,△ABC中,AB=AC,∠A=36°,AB的垂直平分线DE交AC于D,交AB于E,则∠BDC的度数为(  ) A.72° B.36° C.60° D.82°

问题描述:

如图,△ABC中,AB=AC,∠A=36°,AB的垂直平分线DE交AC于D,交AB于E,则∠BDC的度数为(  )
A. 72°
B. 36°
C. 60°
D. 82°

∵AB=AC,∠A=36°,
∴∠ABC=∠C=

180°−∠A
2
=
180°−36°
2
=72°,
∵DE垂直平分AB,
∴∠A=∠ABD=36°,
∴∠BDC=∠A+∠ABD=36°+36°=72°.
故选A.