a. Vì \(\Delta ABC\) vuông tại A
=> \(AB^2+AC^2=BC^2\)
hay \(36^2+48^2=BC^2\)
=> \(\sqrt{BC}=\sqrt{3600}\)
=> BC = 60cm
Xét \(\Delta AHB\) và \(\Delta CAB\) có:
\(\widehat{AHB}=\widehat{CAB}\left(=90^0\right)\)
\(\widehat{B}\left(chung\right)\)
Do đó: \(\Delta AHB\infty\Delta CAB\left(g-g\right)\)
=> \(\dfrac{AH}{AC}=\dfrac{AB}{BC}\)
hay \(\dfrac{AH}{48}=\dfrac{36}{60}\)
=> AH = 28,8 cm
b. Xét \(\Delta CAB\) và \(\Delta CHA\) có:
\(\widehat{CAB}=\widehat{CHA}\left(=90^0\right)\)
\(\widehat{C}\left(chung\right)\)
Do đó: \(\Delta CAB\infty\Delta CHA\left(g-g\right)\)
mà \(\Delta CAB\infty\Delta AHB\left(Cmt\right)\)
=> \(\Delta CHA\infty AHB\)
