Dựng $MH \perp AB (H \in AB)$
Thì ta có $\sin{\alpha}=\frac{MH}{AM} \Rightarrow MH= \sin{\alpha}\cdot AM$
a) Ta có $S_{ABM}=\frac{1}{2} \cdot AB \cdot MH = \frac{1}{2} \cdot AB \cdot AM \cdot \sin{\alpha}$
b) Tương tự phần a), ta có $S_{ACM}= \frac{1}{2} \cdot AC \cdot AM \cdot \sin(90^{\circ}-\alpha) = \frac{1}{2} \cdot AC \cdot AM \cdot \cos{\alpha}$
Có $S_{ABM}+S_{ACM}=S_{ABC}$
$\Leftrightarrow \frac{1}{2} \cdot AB \cdot AM \cdot \sin{\alpha}+ \frac{1}{2} \cdot AC \cdot AM \cdot \cos{\alpha}=\frac{1}{2}AB\cdot AC$
$\Leftrightarrow AM (AB \sin{\alpha}+ AC \cos{\alpha})=AB \cdot AC$
$\Leftrightarrow AM= \frac{AB \cdot AC}{AB \sin{\alpha}+ AC \cos{\alpha}}$