\(a,\) Áp dụng HTL cho tam giác ABK và ACK:
\(\left\{{}\begin{matrix}AE\cdot AB=AK^2\\AF\cdot AC=AK^2\end{matrix}\right.\Rightarrow AE\cdot AB=AF\cdot AC\)
\(b,AE\cdot AB=AF\cdot AC\Rightarrow\dfrac{AE}{AC}=\dfrac{AF}{AB}\)
Mà có \(\widehat{BAC}\) chung
\(\Rightarrow\Delta AEF\sim\Delta ACB\left(c.g.c\right)\)
\(c,AK=\sqrt{AB^2-BK^2}=12\left(cm\right)\left(pytago\right) \)
Áp dụng HTL tam giác: \(AK^2=AB\cdot AE\Rightarrow AE=\dfrac{AK^2}{AB}=\dfrac{12^2}{13}=\dfrac{144}{13}\left(cm\right)\)
Ta có \(KC=BC-BK=13-5=8\left(cm\right)\)
\(AC=\sqrt{AK^2+KC^2}=\sqrt{12^2+8^2}=4\sqrt{13}\left(cm\right)\left(pytago\right)\)
Vì \(\Delta AEF\sim\Delta ACB\) \(\Rightarrow\dfrac{AE}{AC}=\dfrac{EF}{BC}\Rightarrow EF=\dfrac{AE\cdot BC}{AC}=\dfrac{\dfrac{144}{13}\cdot13}{4\sqrt{13}}=\dfrac{144}{4\sqrt{13}}=\dfrac{36\sqrt{13}}{13}\left(cm\right)\)
\(\dfrac{S_{AEF}}{S_{ABC}}=\left(\dfrac{AE}{AC}\right)^2=\left(\dfrac{\dfrac{144}{13}}{4\sqrt{13}}\right)^2=\left(\dfrac{36\sqrt{13}}{169}\right)^2=\dfrac{16848}{28561}=\dfrac{1296}{2197}\)
