Ta có: $∆ABH\sim ∆CAH\, (g.g)$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{AH}{CH}$
$\Rightarrow \dfrac{AB^2}{AC^2} = \dfrac{AH^2}{CH^2} = \dfrac{BH.CH}{CH^2} = \dfrac{BH}{CH}$
$\Rightarrow \dfrac{AB^4}{AC^4} = \dfrac{BH^2}{CH^2} = \dfrac{BM.AB}{CN.AC}$
$\Rightarrow \dfrac{AB^3}{AC^3} = \dfrac{BM}{CN}$
Ta có:
$AB.AC = AH.BC$
$\Rightarrow BC = \dfrac{AB.AC}{AH}$
Ta được:
$BC.BM.CN$
$= \dfrac{AB.AC}{AH}.BM.CN$
$= \dfrac{(AB.BM).(AC.CN)}{AH}$
$= \dfrac{BH^2.CH^2}{AH}$
$= \dfrac{AH^4}{AH}$
$= AH^3 = MN^3$
