Hình vẽ :
a) Xét \(\Delta ABC\) có :
\(\widehat{BAC}=90^o\left(gt\right)\)
\(\Rightarrow BC^2=AB^2+AC^2\) (theo định lí Pitago)
\(\Rightarrow AC^2=BC^2-AB^2\)
\(\Rightarrow AC=\sqrt{BC^2-AB^2}=\sqrt{25^2-15^2}=20\)
b) Xét \(\Delta HAC\) và \(\Delta ABC\) có :
\(\left\{{}\begin{matrix}\widehat{C}:chung\\\widehat{AHC}=\widehat{BAC}=90^o\left(gt\right)\end{matrix}\right.\)
=> \(\Delta HAC\sim\Delta ABC\left(g.g\right)\) (1)
c) Xét \(\Delta ABC,\Delta HBA\) có :
\(\left\{{}\begin{matrix}\widehat{B:}chung\\\widehat{BAC}=\widehat{BHA}=90^o\left(gt\right)\end{matrix}\right.\)
=> \(\Delta ABC\sim\Delta HBA\left(g.g\right)\) (2)
Từ (1) và (2) suy ra : \(\Delta HAC\sim\Delta HBA\left(cùng\sim\Delta ABC\right)\)
=> \(\dfrac{AH}{HB}=\dfrac{HC}{HA}\Rightarrow AH^2=HC.HB\left(đpcm\right)\)
