a) Ta có:
$HM\perp AB; \, HN\perp AC \, (gt)$
$\Rightarrow \widehat{M} = \widehat{N} = 90^o$
$\Rightarrow AMHN$ là hình chữ nhật
$\Rightarrow MN = AH$
Áp dụng định lý Pytago, ta được:
$BC^2 = AB^2 + AC^2$
$= AH^2 + BH^2 + AH^2 + CH^2$
$= BM^2 + MH^2 + CN^2 + NH^2 + 2AH^2$
$= BM^2 + CN^2 + MN^2 + 2AH^2$
$=BM^2 +CN^2 + 3AH^2$
b) Ta có:
$AB.AC = BC.AH = 2S_{ABC}$
$\Rightarrow BC = \dfrac{AB.AC}{AH}$
Ta được:
$BM.CN.BC = BM.CN.\dfrac{AB.AC}{AH}$
$=\dfrac{(AB.BM)(AC.CN)}{AH}$
$=\dfrac{BH^2.CH^2}{AH}$
$=\dfrac{AH^4}{AH} = AH^3$
c) Ta có: $\dfrac{AB}{AC} = \dfrac{AH}{CH}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^2 = \left(\dfrac{AH}{CH}\right)^2$
$\Rightarrow \dfrac{AB^2}{AC^2} = \dfrac{BH.CH}{CH^2} = \dfrac{BH}{CH}$
$\Rightarrow \dfrac{AB^4}{AC^4} = \dfrac{BH^2}{CH^2} = \dfrac{AB.BM}{AC.CN}$
$\Rightarrow \dfrac{AB^3}{AC^3} = \dfrac{BM}{CN}$
