Giải thích các bước giải:
Kẻ $MD\perp AB,D\in AB\to MD//AC$ vì $AB\perp AD$
$\to\sin\alpha=\sin\widehat{DAM}=\dfrac{DM}{AM}$
$\cos\alpha=\cos\widehat{DAM}=\dfrac{AD}{AM}$
$\to \begin{cases}b\cos\alpha=\dfrac{AD\cdot b}{AM}=\dfrac{AD\cdot AC}{AM}\\ c\sin\alpha=\dfrac{DM\cdot c}{AM}=\dfrac{DM\cdot AB}{AM}\end{cases}$
$\to b\cos\alpha+c\sin\alpha=\dfrac{AD\cdot AC+DM\cdot AB}{AM}$
$\to b\cos\alpha+c\sin\alpha=\dfrac{2S_{ADC}+2S_{AMB}}{AM}$
$\to b\cos\alpha+c\sin\alpha=\dfrac{2S_{AMC}+2S_{AMB}}{AM}$ vì $DM//AC\to S_{ADC}=S_{AMC}$
$\to b\cos\alpha+c\sin\alpha=\dfrac{2S_{ABC}}{AM}$
$\to b\cos\alpha+c\sin\alpha=\dfrac{bc}{AM}$
$\to AM=\dfrac{bc}{b\cos\alpha+c\sin\alpha}$
