a)Xét \(\Delta ABC\)và\(\Delta HBA\)
có:góc BAC=góc BHA=90độ
góc B chung
\(\Rightarrow\Delta ABC\)đồng dạng \(\Delta HBA\left(g.g\right)\)
b)xét \(\Delta ABH\)và\(\Delta CAH\) có:
\(\widehat{BAH}=\widehat{ACH}\)(cùng phụ góc CAH)
\(\widehat{ABH}=\widehat{CAH}\)(cùng phụ góc BAH)
\(\Rightarrow\Delta ABH\) đồng dạng \(\Delta CAH\left(g.g\right)\)
\(\Rightarrow\frac{AH}{BH}=\frac{CH}{AH}\Rightarrow AH^2=BH.CH\)