a) Xét \(\Delta ABC,\Delta CAH\) có :
\(\left\{{}\begin{matrix}\widehat{C}:Chung\\\widehat{BAC}=\widehat{AHC}=90^o\end{matrix}\right.\)
=> \(\Delta ABC\sim\Delta CAH\left(g.g\right)\)
Xét \(\Delta ABC,\Delta HBA\) có :
\(\left\{{}\begin{matrix}\widehat{B}:Chung\\\widehat{BAC}=\widehat{BHA}=90^o\end{matrix}\right.\)
=> \(\Delta ABC\sim\Delta HBA\left(g.g\right)\)
=> \(\dfrac{BH}{AB}=\dfrac{AB}{BC}\)
=> \(AB^2=BH.BC\)
b) Ta có: \(AB^2=BH.BC=>AB^2=BH.\sqrt{AB^2+AC^2}\)
=> \(15^2=BH.\sqrt{15^2+20^2}=>BH=\dfrac{15^2}{25}=9\left(cm\right)\)
Từ \(\Delta ABC\sim\Delta CAH\left(g.g\right)\) ta có :
\(\dfrac{AB}{BC}=\dfrac{HC}{AC}=>HC=\dfrac{AB.AC}{BC}=12\left(cm\right)\)
c) Xét \(\Delta MAH,\Delta HAB\) có :
\(\left\{{}\begin{matrix}\widehat{A:}chung\\\widehat{AMH}=\widehat{AHB}=90^o\end{matrix}\right.\)
=> \(\Delta MAH\sim\Delta HAB\left(g.g\right)\)
=> \(\dfrac{MA}{HA}=\dfrac{AH}{AB}=>AH^2=MA.AB\) (1)
Xét \(\Delta NAH,\Delta HAC\) có :
\(\left\{{}\begin{matrix}\widehat{A}:Chung\\\widehat{ANH}=\widehat{AHC}=90^o\end{matrix}\right.\)
=> \(\Delta NAH\sim\Delta HAC\left(g.g\right)\)
=> \(\dfrac{NA}{AH}=\dfrac{AH}{AC}=>AH^2=NA.AC\) (2)
Từ (1) và (2) => \(AM.AB=AN.AC\left(=AH^2\right)\)
d) Xét \(\Delta AMN,\Delta ACB\) có :
\(\dfrac{AM}{AN}=\dfrac{AB}{AC}\)
\(\widehat{A}:Chung\)
=> \(\Delta AMN\sim\Delta ACB\left(g.g\right)\)
