a)\(\widehat{BAH}chung\)
Xét tam giác $AEH$ và $AHB$ có:
\(\left\{\begin{matrix} \widehat{B}-\text{chung}\\ \widehat{AEH}=\widehat{AHB}=90^0\end{matrix}\right.\Rightarrow \triangle AEH\sim \triangle AHB(g.g)\)
\(\Rightarrow \frac{AE}{AH}=\frac{AH}{AB}\Rightarrow AE.AB=AH^2(1)\)
b) Tương tự:
\(\triangle AFH\sim \triangle AHC(g.g)\Rightarrow \frac{AF}{AH}=\frac{AH}{AC}\Rightarrow AF.AC=AH^2(2)\)
c) chịu