a) Xét $∆AHB$ và $∆AKC$ có:
$\widehat{A}:$ góc chung
$\widehat{H} = \widehat{K} = 90^o$
Do đó $∆AHB \sim ∆AKC \, (g.g)$
$\Rightarrow \dfrac{AH}{AK} = \dfrac{AB}{AC}$
$\Rightarrow \dfrac{AH}{AB} = \dfrac{AK}{AC}$
Xét $∆AHK$ và $∆ABC$ có:
$\widehat{BAC}:$ góc chung
$\dfrac{AH}{AB} = \dfrac{AK}{AC}$ $(cmt)$
Do đó $∆AHK\sim ∆ABC \, (c.g.c)$
$\Rightarrow \dfrac{S_{AHK}}{S_{ABC}} = \left(\dfrac{AH}{AB}\right)^2 = \cos^2A$
$\Rightarrow S_{AHK} = S_{ABC}.\cos^2A$
b) Ta có: $S_{BKHC} = S_{ABC} - S_{AHK}$
$= S_{ABC} - S_{ABC}.\cos^2A$
$= S_{ABC}(1 - \cos^2A)$
$= S_{ABC}.\sin^2A$
