Lời giải:
Từ $A$ kẻ $AH\perp BC\ (H\in BC)$
$\Rightarrow\begin{cases}AH\perp MB\\AH\perp MC\end{cases}$
$\Rightarrow \begin{cases}S_{AMB}=\dfrac12MB.AH\\S_{AMC}=\dfrac12MC.AH\end{cases}$
$\Rightarrow \dfrac{S_{AMB}}{S_{AMC}}=\dfrac{MB}{MC}$
Ta lại có: $MB = MC = \dfrac12BC\quad (gt)$
$\Rightarrow \dfrac{MB}{MC}= 1$
$\Rightarrow \dfrac{S_{AMB}}{S_{AMC}}=1$
$\Rightarrow S_{AMB} = S_{AMC}$