- Xét \(\Delta ABH\) và \(\Delta CBA\) có :
\(\left\{{}\begin{matrix}\widehat{BHA}=\widehat{BAC}=90^o\\\widehat{ABH}\left(chung\right)\end{matrix}\right.\)
=> \(\Delta ABH\) ~ \(\Delta CBA\) ( g - g )
=> \(\frac{AB}{AC}=\frac{AH}{AB}\)
=> \(AB^2=AC.AH\)
CMTT : \(AC^2=BC.HC\)
Ta có : \(BH.BC.HC=BH.BC.HC\) ( luôn đúng )
=> \(AB^2.CH=AC^2.BH\) ( đpcm )