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Kẻ đường cao AK của $\Delta$ABC

\RightarrowBK=CK=$\dfrac{BC}{2}$

$\Delta$AKC $\sim$ $\Delta$BHC (g.g)

\Rightarrow$\dfrac{CK}{CH}$=$\dfrac{AC}{BC}$ \Leftrightarrow $\dfrac{BC}{2CH}$=$\dfrac{AC}{BC}$ (vì CK=$\dfrac{1}{2}BC$)

\LeftrightarrowAC=$\dfrac{BC^2}{2CH}$

\Leftrightarrow$\dfrac{AC}{2}$=$\dfrac{BC^2}{4CH}$=$\dfrac{BH.BC^2}{4BH.CH}$

=$\dfrac{BH}{4}$.$\dfrac{BC}{BH}.\dfrac{BC}{CH}$=$\dfrac{BH}{4sin\alpha cos\alpha}$

\Leftrightarrow$\dfrac{BH^2}{4sin\alpha cos\alpha}$=$\dfrac{BH.AC}{2}$=$S_{ABC}$
 
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