Ta có: \(\dfrac{AB}{BC}=\dfrac{3}{5}\)
nên \(AB=\dfrac{3}{5}BC\)
Ta có: \(AB^2=BH\cdot BC\)
\(\Leftrightarrow\dfrac{9}{25}BC^2-a\cdot BC=0\)
\(\Leftrightarrow BC\cdot\left(\dfrac{9}{25}BC-a\right)=0\)
\(\Leftrightarrow BC\cdot\dfrac{9}{25}=a\)
hay \(BC=a:\dfrac{9}{25}=\dfrac{25}{9}a\)
\(\Leftrightarrow AB=\dfrac{3}{5}BC=\dfrac{3}{5}\cdot\dfrac{25}{9}a=\dfrac{5}{3}a\)
\(\Leftrightarrow CH=BC-BH=\dfrac{25}{9}a-a=\dfrac{16}{9}a\)
\(\Leftrightarrow AC=\sqrt{\left(\dfrac{25}{9}a\right)^2-\left(\dfrac{5}{3}a\right)^2}=\dfrac{20}{9}a\)
\(\Leftrightarrow AH=\sqrt{\left(\dfrac{20}{9}a\right)^2-\left(\dfrac{16}{9}a\right)^2}=\dfrac{4}{3}a\)
