Cho tam giác ABC vuông tại A, AB cho tam giác ABC vuông tại A,AB<AC.phân giác AD.AJ vuông góc BC.DH vuông góc AB,DK vuông góc AC.BK cắt DH tại M,CHcắt DK tại N. a)chứng minh rằng HM/MD=BH/BA b)MN song song BC c)gọi I là giao điểm BK và CH. chứng minh tam giác ABK đồng dạng với tam giác KAN và 3 điểm A,I,J thẳng hàng
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31 Tham khảo câu trả lời từ Trợ lý ảo Lazi a) Ta có: $\frac{HM}{MD} = \frac{HM}{BK} \cdot \frac{BK}{MD} = \frac{CH}{DK} \cdot \frac{DH}{BA} = \frac{BH}{BA}$ Vậy ta có $HM/MD=BH/BA$ b) Ta có $\angle HMD = \angle BAC = \angle BDC$ (do $ABCD$ là tứ giác nội tiếp), từ đó suy ra $MN$ song song với $BC$. c) Ta có $\angle BAK = \angle BAC = \angle BDC = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = \angle BDK = 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