a) Ta có: $\widehat{A} = \widehat{E} = \widehat{F} = 90^o$
$\Rightarrow AEHF$ là hình chữ nhật
$\Rightarrow \widehat{AEF} = \widehat{HAE}$
mà $\widehat{HAE} = \widehat{HCA}$ (cùng phụ $\widehat{HAC}$)
nên $\widehat{AEF} = \widehat{HCA}$
hay $\widehat{AEF} = \widehat{BCA}$
Xét $ΔAEF$ và $ΔACB$ có:
$\widehat{A}:$ góc chung
$\widehat{AEF} = \widehat{BCA}$ $(cmt)$
Do đó $ΔAEF\sim ΔACB \, (g.g)$
b) Áp dụng định lý Pytago, ta được:
$BC^2 = AB^2 + AC^2$
$=BH^2 + AH^2 + CH^2 + AH^2$
$= BE^2 + EH^2 + AH^2 + CF^2 + HF^2 + AH^2$
$=BE^2 + CF^2 + (EH^2 + HF^2) + 2AH^2$
$= BE^2 + CF^2 + EF^2 + 2AH^2$
$=BE^2 + CF^2 + AH^2 + 2AH^2$ ($EF = AH: \, AEHF$ là hình chữ nhật)
$= BE^2 + CF^2 + 3AH^2$
c) Ta có: $ABH \sim ΔCAH$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{BH}{AH}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^2 = \left(\dfrac{BH}{AH}\right)^2 = \dfrac{BH^2}{BH.CH} = \dfrac{BH}{CH}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^4 = \left(\dfrac{BH}{CH}\right)^2 = \dfrac{BE.AB}{CF.AC}$
$\Rightarrow \left(\dfrac{AB}{AC}\right)^3 = \dfrac{BE}{CF}$
d) Ta có: $AB.AC = BC.AH = 2S_{ABC}$
$\Rightarrow BC = \dfrac{AB.AC}{AH}$
Ta được: $BC.BE.CF = \dfrac{AB.AC}{AH}.BE.CF$
$=\dfrac{(AB.BE).(AC.CF)}{AH}$
$=\dfrac{BH^2.CH^2}{AH}$
$=\dfrac{AH^4}{AH} = AH^3$
