Distance from centroid to hypotenuse We got triangle $\triangle \text{ABC}$ and $\angle \text{C}=90^{\circ }$. The area of $\triangle \text{ABC}=\text{S}$ and $\angle \text{BAC}=\alpha$. Prove that the distance from the centroid to the hypotenuse is equal to $\frac{1}{2}\sqrt{\text{S}\sin\left(2\alpha \right)}$.
Any ideas how to solve that? Would be grateful if someone helps me.
 A: Remember the theorems "In a right triangle, the length of the median to the hypotenuse equals half the hypotenuse", and also "The centroid in a triangle is a point which divides each median in a proportion 2:1, where the longest side is always between the centroid and its vertex.
So we have

Call the centroid $\;M\;$ and let $\;K\;$ be the midpoint of $\;AB\;$, and let $\;x=|AB|=$ the hypotenuse's length . Thus, we have:
$$\cos\alpha=\frac{AC}x\implies AC=x\cos\alpha\implies S=\frac{AC\cdot AB\cdot\sin\alpha}2=\frac{x^2\cos\alpha\sin\alpha}2=\frac14x^2\sin2\alpha$$
Now, observe that summing angles of triangle $\;\Delta BCK\;$ , we have that 
$$\;\angle BKC=2\alpha\implies\angle MKA=180^\circ-2\alpha\;$$
Also
$$|CK|=\frac12x\implies |MK|=\frac13|CK|=\frac x6$$
so we can know form a little right triangle $\;\Delta MKL\;$ , with $\;L\;$ a point on the hypotenuse $\;AB\;$ s.t. $\;ML\perp AB\;$ , and then in this right triangle:
$$\sin(180-2\alpha)=\frac{ML}{MK}\implies ML=MK\sin2\alpha=\frac {x\sin2\alpha}6$$
Finally, observe that 
$$\frac12\sqrt{S\sin2\alpha}=\frac12\sqrt{\frac14x^2\sin^22\alpha}=\frac14 x\sin2\alpha$$
You can see I get a different result, so either I am wrong (check the above ) or else it should be $\;\frac13\;$ instead of $\;\frac12\;$ in your formula...
A: Let $h$ is an altitude to $AB$.
Thus, $AC=\frac{h}{\sin\alpha}$ and $BC=\frac{h}{\cos\alpha}$, which says that
$$S=\frac{h^2}{2\sin\alpha\cos\alpha}=\frac{h^2}{\sin2\alpha},$$
which gives that our distance is $\frac{1}{3}h=\frac{1}{3}\sqrt{S\sin2\alpha}$.
