\(HB.HC=15^2=225\)
Ta có : \(\hept{\begin{cases}AB^2=BH.BC\\AC^2=CH.BH\end{cases}\Rightarrow\frac{AB^2}{AC^2}=\frac{BH}{CH}\Rightarrow\hept{\begin{cases}\frac{HB}{HC}=\frac{25}{49}\\HB.HC=225\end{cases}\Rightarrow}\hept{\begin{cases}HB.HC.\frac{HB}{HC}=\frac{25}{49}.225\\HB.HC=225\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}HB^2=\frac{5625}{49}\\HB.HC=225\end{cases}\Rightarrow\hept{\begin{cases}HB=\frac{75}{7}\\HC=21\end{cases}}}\)
Đặt AB=5a,AC=7a Khi đó, áp dụng HTL ta có
\(\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{AH^2}\)
\(\Leftrightarrow\frac{1}{25a^2}+\frac{1}{49a^2}=\frac{1}{225}\)
\(\Leftrightarrow a=\frac{3\sqrt{74}}{7}\)
Vậy \(AB=\frac{15\sqrt{74}}{7},AC=3\sqrt{74}\)
Áp dụng HTL ta có
AB.AC=AH.BC
\(\Leftrightarrow BC=\frac{222}{7}\)
Áp dụng HTL ta có
\(AB^2=BH.BC\)
\(\Leftrightarrow BH=\frac{AB^2}{BC}=\frac{75}{7}\)
Vậy CH=BC−BH=21
