Cho\(\Delta ABC\) co AB=AC va M la trung diem cua BC.Tren tia doi cua BC lay diem D,tren tia doi cua CB lay diem E sao cho BD=CE.Ke \(BK\perp AD\left(K\in AD\right)\).Tren tia doi cua BK lay diem H sao cho BH=AE.Tren tia doi cua AM lay diem N sao cho AN=CE.Chung minh \(\widehat{MAD}=\widehat{MBH}\) va \(DN\perp DH\)
góc MAD+góc MDA=90 độ
góc MBH=góc KBD=90 độ-góc MDA
=>góc MAD=góc MBH
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