\(a,\Delta ABD\perp A\Rightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{AB^2+BC^2}=17\left(cm\right)\left(pytago\right)\)
\(b,\) Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}AB^2=HB\cdot BD\\AD^2=HD\cdot BD\\AH^2=BH\cdot HD\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}HB=\dfrac{AB^2}{BD}=\dfrac{64}{17}\left(cm\right)\\HD=\dfrac{AD^2}{BD}=\dfrac{225}{17}\left(cm\right)\\AH=\sqrt{\dfrac{64}{17}\cdot\dfrac{225}{17}}=\dfrac{120}{17}\left(cm\right)\end{matrix}\right.\)
\(c,\) Áp dụng HTL cho tam giác AHB,AHD và ABD
\(\left\{{}\begin{matrix}EA\cdot EB=EH^2\\FA\cdot FD=FH^2\\HB\cdot HD=HA^2\left(1\right)\end{matrix}\right.\Rightarrow EA\cdot EB+FA\cdot FD=EH^2+FH^2\left(2\right)\)
Ta có \(\widehat{EAF}=\widehat{AFH}=\widehat{AEH}=90^0\) nên AEHF là hcn
\(\Rightarrow HF=AE\Rightarrow HF^2=AE^2\left(3\right)\\ \left(2\right)\left(3\right)\Rightarrow EA\cdot EB+FA\cdot FD=EH^2+AE^2=AH^2\left(định.lí.pytago.\Delta AEH\right)\left(4\right)\\ \left(1\right)\left(4\right)\RightarrowĐpcm\)
