Legs from the altitude and the ratio of the legs

Text

In triangle $A\hat{C}B$, right-angled at $\hat{C}$, the altitude $\overline{CH}$ to the hypotenuse equals $2$. Moreover one leg is $\dfrac43$ of the other. Determine all the sides of the triangle.

Solution

Let the legs be $\overline{BC}=b$ and $\overline{AC}=\dfrac43 b$. By the Pythagorean theorem the hypotenuse is $\overline{AB}=\sqrt{b^2+\tfrac{16}{9}b^2}=\sqrt{\tfrac{25}{9}b^2}=\frac53 b.$ The altitude to the hypotenuse follows from the area (computed two ways): $\;\overline{CH}=\dfrac{\overline{AC}\cdot\overline{BC}}{\overline{AB}}$: $2=\frac{\frac43 b\cdot b}{\frac53 b}=\frac45\,b\ \Longrightarrow\ b=\frac{5}{2}=2{,}5.$ Hence $\overline{BC}=2{,}5,\qquad \overline{AC}=\frac{10}{3}\approx3{,}33,\qquad \overline{AB}=\frac{25}{6}\approx4{,}17.$ $\boxed{\overline{BC}=\tfrac52,\quad \overline{AC}=\tfrac{10}{3},\quad \overline{AB}=\tfrac{25}{6}.}$