a: Kẻ đường cao AH
\(S_{AMB}=\dfrac{1}{2}\cdot AH\cdot MB\)
\(S_{AMC}=\dfrac{1}{2}\cdot AH\cdot MC\)
mà MB=MC
nên \(S_{AMB}=S_{AMC}\)
MB=1/2BC
=>\(S_{AMB}=\dfrac{1}{2}\cdot AH\cdot\dfrac{1}{2}\cdot BC=\dfrac{1}{2}\cdot S_{ABC}\)
=>ĐPCM
b: Xét ΔABC có BC^2=AB^2+AC^2
nên ΔABC vuông tại A
AN=AC/2=4cm
\(S_{ANB}=\dfrac{1}{2}\cdot4\cdot6=2\cdot6=12\left(cm^2\right)\)
\(S_{ABC}=\dfrac{1}{2}\cdot6\cdot8=24\left(cm^2\right)\)
=>\(S_{BNC}=12\left(cm^2\right)\)
=>\(S_{BMN}=\dfrac{1}{2}\cdot12=6\left(cm^2\right)\)
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