a) Xét $∆AEB$ và $∆AFC$ có:
$\widehat{AEB} = \widehat{AFC} = 90^o$
$\widehat{A}:$ góc chung
Do đó $∆AEB\sim ∆AFC\, (g.g)$
b) Ta có: $∆AEB\sim ∆AFC$
$\Rightarrow \dfrac{AE}{AF} = \dfrac{AB}{AC}$
$\Rightarrow \dfrac{AE}{AB} = \dfrac{AF}{AC}$
Xét $∆AEF$ và $∆ABC$ có:
$\dfrac{AE}{AB} = \dfrac{AF}{AC}$
$\widehat{A}:$ góc chung
Do đó $∆AEF\sim ∆ABC \, (c.g.c)$
$\Rightarrow \dfrac{S_{AEF}}{S_{ABC}} = \left(\dfrac{AE}{BC}\right)^2 = \left(\dfrac{3}{6}\right)^2 = \dfrac{1}{4}$
$\Rightarrow S_{ABC} = 4S_{AEF}$