Xét tam giác AEF và tam giác ABC có:
A chung
\(\dfrac{AE}{AB}=\dfrac{AF}{AC}\left(=cosA\right)\)
\(\Rightarrow\Delta AEF\sim\Delta ABC\left(c.g.c\right)\)
\(\Rightarrow\dfrac{S_{AEF}}{S_{ABC}}=\left(\dfrac{AE}{AB}\right)^2=cos^2A=1-sin^2A\)
\(1-\sin^2A=\cos^2A=\dfrac{AF^2}{AC^2}\left(1\right)\)
Ta có \(\widehat{AEB}=\widehat{AFC}=90^0\Rightarrow\Delta AEB\sim\Delta AFC\left(g.g\right)\)
\(\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\Rightarrow\Delta AEF\sim\Delta ABC\left(c.g.c\right)\\ \Rightarrow\dfrac{S_{AEF}}{S_{ABC}}=\left(\dfrac{AF}{AC}\right)^2=\dfrac{AF^2}{AC^2}\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
