a) Xét \(\Delta ABC,\Delta HBA\) có :
\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{BHA}\left(=90^o\right)\\\widehat{B}:Chung\end{matrix}\right.\)
\(\Rightarrow\)\(\Delta ABC\sim\Delta HBA\left(g.g\right)\)
b) Ta có : \(S_{\Delta ABC}=\)\(\left\{{}\begin{matrix}\dfrac{1}{2}AB.AC\\\dfrac{1}{2}AH.BC\end{matrix}\right.\Rightarrow AB.AC=AH.BC\)
\(\Rightarrow8.15=AH.17\)
\(\Rightarrow AH=\dfrac{8.15}{17}=\dfrac{120}{17}\left(cm\right)\)
c) Xét \(\Delta ABH,\Delta AHM\) có :
\(\left\{{}\begin{matrix}\widehat{A}:Chung\\\widehat{AHB}=\widehat{AMH}=90^o\end{matrix}\right.\)
=> \(\Delta ABH\sim\Delta AHM\left(g.g\right)\)
=> \(\dfrac{AH}{AB}=\dfrac{AM}{AH}\)
=> \(AH^2=AM.AB\left(1\right)\)
Xét \(\Delta ACH,\Delta AHN\) có :
\(\left\{{}\begin{matrix}\widehat{A}:chung\\\widehat{AHC}=\widehat{ANH}=90^o\end{matrix}\right.\)
=> \(\Delta ACH\sim\Delta AHN\left(g.g\right)\)
=> \(\dfrac{AH}{AC}=\dfrac{AN}{AH}\)
=> \(AH^2=AN.AC\left(2\right)\)
Từ (1) và (2) => \(AM.AB=AN.AC\left(=AH^2\right)\)
=> đpcm.