Legs of a right triangle via Pythagoras and Euclid

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In a right triangle $A\hat{B}C$ (right angle at $B$) the hypotenuse is $\overline{AC}=9$. Knowing that $\overline{BH}^2+\overline{AB}^2=5\,\overline{AH}^2$, where $\overline{BH}$ is the altitude to the hypotenuse and $H$ its foot, find the legs using the theorems of Pythagoras and Euclid.

Solution

Let $\overline{AH}=p$ and $\overline{HC}=q$, with $p+q=9$. Euclid’s theorems give: $\overline{AB}^2 = p\cdot \overline{AC}=9p, \qquad \overline{BH}^2 = p\cdot q.$ Substitute into the condition $\overline{BH}^2+\overline{AB}^2=5\,\overline{AH}^2$: $pq + 9p = 5p^2 \implies q + 9 = 5p \implies q = 5p-9.$ With $p+q=9$: $p+(5p-9)=9 \implies 6p=18 \implies p=3$, hence $q=6$. Therefore $\overline{AB}^2=9\cdot 3=27 \implies \overline{AB}=3\sqrt3\approx 5{,}196,$ $\overline{BC}^2=q\cdot \overline{AC}=6\cdot 9=54 \implies \overline{BC}=3\sqrt6\approx 7{,}348.$

$\boxed{\overline{AB}=3\sqrt3,\qquad \overline{BC}=3\sqrt6}$