Xét $ΔABC$ vuông tại $A$ đường cao $AH$ có: $AH^2=BH.HC=6^2=36$ $⇒HB=\dfrac{36}{HC}$ Thay vào $\dfrac{BH}{HC}=\dfrac{4}{9}$ ta được: $\dfrac{\dfrac{36}{HC}}{HC}=\dfrac{4}{9}$ $⇔\dfrac{36}{HC^2}=\dfrac{4}{9}$ $⇔4HC^2=324$ $⇔HC^2=81$ $⇔HC=9$ hoặc $HC=-9(L)(cm)$ $⇒HB=\dfrac{36}{HC}=\dfrac{36}{9}=4(cm)$
$⇒BC=BH+HC=4+9=13(cm)$ $AB^2=BH.BC$ (Hệ thức lượng) $⇒AB=\sqrt{BH.BC}=\sqrt{4.13}=2\sqrt{13}(cm)$ $AC^2=HC.BC$ $⇒AC=\sqrt{HC.BC}=\sqrt{9.13}=3\sqrt{13}(cm)$
