a: ΔHBA\(\sim\)ΔABC vì \(\widehat{B}\) chung; góc AHB=góc CAB=90 độ
ΔHCA\(\sim\)ΔACB vì góc C chung, góc AHC=góc BAC=90 độ
b: \(BC=\sqrt{a^2+b^2}\)
\(BH=\dfrac{AB^2}{BC}=\dfrac{a^2}{\sqrt{a^2+b^2}}\)
\(CH=\dfrac{AC^2}{BC}=\dfrac{b^2}{\sqrt{a^2+b^2}}\)
\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{a\cdot b}{\sqrt{a^2+b^2}}\)