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a) M điểm giữa của BC nên CM = MB
Tam giác ABM và tam giác ACM có : đáy CM = MB, chung đường cao tương ứng với đáy.
Nên S(ABM) = S(ACM)
b) S(ABC) = S(ABM) + S(ACM) mà S(ABM) = S(ACM)
Nên \(S\left(ABM\right)=\frac{1}{2}S\left(ABC\right)\)
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