题目内容
如图,在△ABC中,AB=AC,∠A=52°,AB的垂直平分线MN交AC于点D.求∠DBC的度数.
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∵AB=AC,∠A=52°,∴∠ABC=∠ACB=
=64°,
∵AB的垂直平分线MN,∴AD=BD,∠A=∠ABD=52°,
∴∠DBC=∠ABC-∠ABD=64°-52°=12°.
180°-52° |
2 |
∵AB的垂直平分线MN,∴AD=BD,∠A=∠ABD=52°,
∴∠DBC=∠ABC-∠ABD=64°-52°=12°.
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