解已知,∠A=50°,AB=AC⇒∠ABC=∠ACB=65°
又∵DE垂直且平分AB⇒DB=AD
∴∠ABD=∠A=50°
∴∠DBC=∠ABC-∠ABD=65°-50°=15°.
故选A.
如图,在△ABC中,∠A=50°,AB=AC,AB的垂直平分线DE交AC于D,则∠DBC的度数是( ) A.15° B.20° C.30° D.25°
如图,在△ABC中,∠A=50°,AB=AC,AB的垂直平分线DE交AC于D,则∠DBC的度数是( )
A. 15°
B. 20°
C. 30°
D. 25°
A. 15°
B. 20°
C. 30°
D. 25°
数学人气:782 ℃时间:2020-01-30 11:27:37
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