Cho hình tam giác ABC ;M là trung điểm của BC.Nối A với M.Hãy chứng tỏ rằng diện tích hai hình tam giác ABM và ACM bằng nhau.
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
Ta có: \(S_{ABD}-S_{ABC}=13,2\)
=>\(S_{ACD}=13,2\)
Vì \(\frac{BC}{CD}=\frac{32}{8}=4\)
nên \(\frac{S_{ABC}}{S_{ACD}}=4\)
=>\(S_{ABC}=13,2\times4=52,8\left(\operatorname{cm}^2\right)\)
Vì BM = MC mà 2 hình đều có chung chiều cao AM
=> Diện tích 2 hình bằng nhau
chứng minh theo trường hợp bao la