Looking for multiple solutions to show my students. Let ABC be a right triangle with B the right angle. X,Y and Z are on BC, CA and AB respectively such that BXYZ is a square. If the square is of side length m, AY = r and YC = s, find m in terms of r and s.
I have two solutions for my students (I instruct a math team class): One involves proportions and the other involves the angle bisector theorem (BY is an angle bisector). I am just curious of there are any other creative solutions for this problem.
 A: Let $\theta=\angle ZYA=\angle XCY$. Then $\cos\theta=\frac{m}{r}$ and $\sin\theta=\frac{m}{s}$. It follows that $\left(\frac{m}{r}\right)^2+\left(\frac{m}{s}\right)^2=1$. 
A: You described this in the comments, so perhaps you have it already, but circumscribe the circle through $\rm ABC$ with center at $\rm O$ (on $\rm AC$ since $\angle\rm B$ is a right angle) and extend $\rm BY$ to meet the circle again at $\rm M$.

Then $\angle \mathrm{AOM}=2\angle\mathrm{ABM}$ is a right angle. Wlog I assume $\rm AB > BC$ and $s>r$. So $\mathrm{OM}=\mathrm{AO}=(s+r)/2$, $\mathrm{OY}=(s-r)/2$. Then by the Pythagorean theorem $|\mathrm{YM}|^2=(s^2+r^2)/2$, and by the intersecting chords theorem
$$
\mathrm{BY}\cdot\mathrm{YM} = \mathrm{CY}\cdot\mathrm{YA} \\
\sqrt{2}m\cdot \sqrt{(s^2+r^2)/2} = r\cdot s \\
m = \frac{rs}{\sqrt{s^2+r^2}}
$$
A: The solution for someone who really prefers algebra to geometry:
Let $AZ=p$, $XC=q$. Then the Pythagorean theorem tells us that
\begin{eqnarray}
p^2+m^2&=&r^2\\
q^2+m^2&=&s^2\\
(p+m)^2+(q+m)^2&=&(r+s)^2 \, .
\end{eqnarray}
Subtracting the first two equations from the third and simplifying yields
$p+q=\frac{rs}{m}$, while subtracting the second equation from the first yields $p^2-q^2=r^2-s^2$. So $p-q=\frac{m}{rs}(r^2-s^2)$, and thus 
$$p=\frac{1}{2}\left(\frac{mr}{s}+\frac{rs}{m}-\frac{ms}{r}\right) \, .$$
Substituting this value into the first Pythagorean relation, we have
$$\frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}+2r^2-2s^2-2m^2\right)+m^2-r^2=0 \\
\frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}-2r^2-2s^2+2m^2\right)=0 \\
\left(\frac{mr}{s}-\frac{rs}{m}+\frac{ms}{r}\right)^2=0 \, ;$$
isolating $m$ gives
$$m^2=\frac{rs}{\frac{r}{s}+\frac{s}{r}} \, .$$
A: Rotate the triangle $XYC$ by $90$ degrees about $Y$ in such a way as to make $XY$ coincide with $YZ$. This gives a new right triangle; its legs have length $r$ and $s$, and the altitude to its hypotenuse has length $m$. So the area of this triangle is $\frac{1}{2}rs=\frac{1}{2}m\sqrt{r^2+s^2}$.
