a) Ta có: $\cos B = \dfrac{BH}{AB}$
$\Rightarrow AB\cos B = BH$
$\cos C = \dfrac{CH}{AC}$
$\Rightarrow AC\cos C = CH$
$\Rightarrow AB\cos B + AC\cos C = BH + CH = BC$
b) Ta có: $AEHF$ là hình chữ nhật
$\Rightarrow S_{AEHF} = HE.HF = \dfrac{AH.BH}{AB}\cdot\dfrac{AH.CH}{AC}$
$= \dfrac{AH^2.BH.CH}{AB.AC} = \dfrac{AH^4}{BC.AH} = \dfrac{AH^3}{BC}$
c) Ta có: $AB.AC = BC.AH$
$\Rightarrow BC = \dfrac{AB.AC}{AH}$
Ta được:
$EB.BC.CF = \dfrac{AB.AC}{AH}.BE.CF$
$= \dfrac{AB.BE.AC.CF}{AH} = \dfrac{BH^2.CH^2}{AH} = \dfrac{AH^4}{AH} = AH^3$
d) Ta có: $AH^3 = EB.BC.CF$ (câu c)
$\Rightarrow BE.CF = \dfrac{AH^3}{BC}$
Ta được:
$\dfrac{AH^2}{BE.CF}$
$= \dfrac{AH^2}{\dfrac{AH^3}{BC}} = \dfrac{BC}{AH}$
$= \dfrac{BC^2}{BC.AH} = \dfrac{BC^2}{AB.AC} = \dfrac{AB^2 + AC^2}{AB.AC} = \dfrac{AC}{AB} + \dfrac{AB}{AC}$
