Ta có:
$ΔABI \sim ΔACK \, (g.g)$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{AI}{AK}$
$\Rightarrow \dfrac{AI}{AB} = \dfrac{AK}{AC}$
Xét $ΔAIK$ và $ΔABC$ có:
$\widehat{A}:$ góc chung
$\dfrac{AI}{AB} = \dfrac{AK}{AC} \quad (cmt)$
Do đó $ΔAIK\sim ΔABC\, (c.g.c)$
$\Rightarrow \dfrac{S_{AIK}}{S_{ABC}} = \left(\dfrac{AI}{AB}\right)^2 = \cos^2A$
$\Rightarrow S_{AIK} = \cos^2A.S_{ABC}$
Chứng minh tương tự, ta được:
$S_{BHK} = \cos^2B.S_{ABC}$
$S_{CHI} = \cos^2C.S_{ABC}$
$\Rightarrow S_{HIK} = S_{ABC} - S_{AIK} - S_{BHK} - S_{CHI}$
$= S_{ABC} - \cos^2A.S_{ABC} - \cos^2B.S_{ABC} - \cos^2C.S_{ABC}$
$\Rightarrow S_{HIK} = S_{ABC}(1 - \cos^2A - \cos^2B - \cos^2C)$
$\Rightarrow \dfrac{S_{HIK}}{S_{ABC}} = 1 - \cos^2A - \cos^2B - \cos^2C$
