Giải thích các bước giải:
a)
Xét `\DeltaABC`, ta có:
\(\left \{ {{\hat{A}=\hat{AHB}} \atop {\hat{B}\ chung}} \right.\)
`=>ΔABC\ ~\ ΔHBA\ (g.g)`
b)
Ta có: `BC=\sqrt{AB^2+AC^2}=\sqrt{20^2+15^2}=\sqrt{625}=25cm`
Ta có: `ΔABC\ ~\ ΔHBA\ (cmt)`
`=>(AB)/(BH)=(BC)/(AB)`
`=>15/(BH)=25/15`
`=>BH=(15.15)/25=9cm`
c)
Ta có: `\hat{AHB}=\hat{AHC}`
`=>ΔAHB\ ~\ ΔCHA\ (g.g)`
`=>(AH)/(HC)=(HB)/(AH)`
`=>AH^2=HB.HC`