$P_{HAB}^2= (AB + BH + AH)^2 = AB^2 + BH^2 + AH^2 + 2AB.AH + 2AB.BH + 2AH.BH$
$P_{HAC}^2 = (AC + AH + CH)^2 = AC^2 + AH^2 + CH^2 + 2AC.AH + 2AC.CH + 2AH.CH$
$\Rightarrow P_{HAB}^2 + P_{HAC}^2$
$= (AB^2 + AC^2 + AH^2 + BH^2 + AH^2 + CH^2) + (2AB.AH + 2.AC.CH) + (2AB.BH+2AC.AH) + (2BH.AH + 2CH.AH)$
Do $∆ABH \sim ∆CAH$
nên $\dfrac{AB}{AC} = \dfrac{BH}{AH} = \dfrac{AH}{CH}$
và $AH.BC = AB.AC = 2S_{ABC}$
$\Rightarrow \begin{cases}AB.AH = AC.BH\\AC.AH = AB.CH\\AH.CH + AH.BH = AB.AC\end{cases}$
Ta được:
$P_{HAB}^2 + P_{HAC}^2$
$= BC^2 + AB^2 + AC^2 + 2(AC.BH + AC.CH) + 2(AB.BH + AB.CH) + 2AB.AC$
$= AB^2 + AC^2 + BC^2 + 2AC.BC + 2AB.BC + 2AB.AC$
$= (AB + AC + BC)^2 = P_{ABC}^2$
