Ta có $\dfrac{AB}{AC}=\dfrac{5}{7}⇒AB=5x,AC=7x$
Xét $ΔABC$ vuông tại $A$
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$ ( hệ thức lượng )
$\dfrac{1}{15^2}=\dfrac{1}{(5x)^2}+\dfrac{1}{(7x)^2}$
$\dfrac{1}{225^2}=\dfrac{1}{25x^2}+\dfrac{1}{49x^2}$
$⇒x=\dfrac{3\sqrt[]{74}}{7}$
$⇒AB=5.\dfrac{3\sqrt[]{74}}{7}=\dfrac{15\sqrt[]{74}}{7}$
$AC=7.\dfrac{3\sqrt[]{74}}{7}=3\sqrt[]{74}$
$⇒BC^2=AB^2+AC^2$
$⇒BC=\sqrt[]{(\dfrac{15\sqrt[]{74}}{7})^2+(3\sqrt[]{74})^2}$
$⇒BC=\dfrac{222}{7}$
$AB^2=BH.BC$
$(\dfrac{15\sqrt[]{74}}{7})^2=BH.\dfrac{222}{7}$
$BH=\dfrac{75}{7}$
$⇒CH=BC-BH=\dfrac{222}{7}-\dfrac{75}{7}=21$
