Sides of the triangle from a projection and the area of a square

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A right triangle $A\hat{C}B$, right-angled at $\hat{C}$, has the altitude $CH$ to the hypotenuse dividing $AB$ into two parts; $\overline{HB}=3$ is known. Determine the legs and the hypotenuse that make the area of the square on leg $BC$ equal to $12$.

Solution

The square on $BC$ has area $\overline{BC}^2=12$. By Euclid’s first theorem applied to leg $BC$ (with projection $HB$ on the hypotenuse): $\overline{BC}^2=\overline{AB}\cdot\overline{HB}\ \Longrightarrow\ 12=3\,\overline{AB}\ \Longrightarrow\ \overline{AB}=4.$ Then $\overline{AH}=4-3=1$. Again by the first theorem, for leg $AC$: $\overline{AC}^2=\overline{AB}\cdot\overline{AH}=4\ \Longrightarrow\ \overline{AC}=2.$ By Euclid’s second theorem the altitude: $\overline{CH}^2=\overline{AH}\cdot\overline{HB}=3\Rightarrow \overline{CH}=\sqrt3$. Finally $\overline{BC}=2\sqrt3$. $\boxed{\overline{BC}=2\sqrt3\approx3{,}46,\quad \overline{AC}=2,\quad \overline{AB}=4,\quad \overline{CH}=\sqrt3\approx1{,}73.}$