\(a,\) Áp dung HTL vào tg ABH vuông tại H
\(HE^2=AE\cdot EB=23,04\Rightarrow HE=4,8\left(cm\right)\)
Do đó \(S_{ABH}=\dfrac{1}{2}HE\cdot AB=\dfrac{1}{2}\cdot4,8\cdot\left(6,4+3,6\right)=24\left(cm^2\right)\)
\(b,\) Áp dụng HTL ta có \(\left\{{}\begin{matrix}AE\cdot AB=AH^2\\AF\cdot AC=AH^2\end{matrix}\right.\Rightarrow AE\cdot AB=AF\cdot AC\Rightarrow\dfrac{AE}{AC}=\dfrac{AF}{AB}\)
Xét \(\Delta AEF\) và \(\Delta ACB\) có
\(\left\{{}\begin{matrix}\dfrac{AE}{AC}=\dfrac{AF}{AB}\\\widehat{BAC}.chung\end{matrix}\right.\) nên \(\Delta AEF\sim\Delta ACB\left(c.g.c\right)\)
\(c,\sin^2C=\dfrac{AH^2}{AC^2}\)
Mà \(\dfrac{AE}{AC}=\dfrac{AF}{AB}\Rightarrow AF=\dfrac{AE\cdot AB}{AC}\Rightarrow\dfrac{AF}{AC}=\dfrac{AE\cdot AB}{AC^2}=\dfrac{AH^2}{AC^2}\)
Do đó \(\dfrac{AF}{AC}=\sin^2C\)
\(d,\sin^2B\cdot\sin^2C=\dfrac{AH^2}{AB^2}\cdot\dfrac{AH^2}{AC^2}=\dfrac{AE\cdot AB}{AB^2}\cdot\dfrac{AF\cdot AC}{AC^2}=\dfrac{AE}{AB}\cdot\dfrac{AF}{AC}=\dfrac{AE}{AB}\cdot\dfrac{AE}{AB}=\dfrac{AE^2}{AB^2}\left(\dfrac{AE}{AB}=\dfrac{AF}{AC}\right)\)
Mà \(\Delta AEF\sim\Delta ACB\) nên \(\dfrac{S_{AEF}}{S_{ABC}}=\dfrac{AE^2}{AB^2}\)
Từ đó ta được đpcm
