来自龙翀的问题
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°
1回答
2020-06-16 19:16
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°
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.