Đặt $AB=5x(cm); AC=7x(cm)$ ($x>0$)
Theo hệ thức lượng:
$\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}$
$\to \dfrac{1}{25x^2}+\dfrac{1}{49x^2}=\dfrac{1}{15^2}$
$\to x=\dfrac{3\sqrt{74}}{7}$
$\to AB=\dfrac{15\sqrt{74}}{7}(cm); AC=3\sqrt{74}(cm)$
Theo Pytago:
$HB=\sqrt{AB^2-AH^2}=\dfrac{75}{7}(cm)$
$HC=\sqrt{AC^2-AH^2}=21(cm)$