Geometry Ratio in right triangle 
In ABC, $\angle ACB=90$. Point O is chosen inside ABC such that [AOB]=[BOC]=[COA]. If OA=23 and OB=11, find $OC^2$.

Using the areas, we can find OD:OE:OF, where D, E, F are the feet of the altitudes from O to BC, AC, and AB, respectively. I have no idea how to proceed

With $D$ and $E$ as stated, note that $OD=\frac{AC}3$, $OE=\frac{BC}3$, then $AE=\frac{2AC}3$, $BD=\frac{2BC}3$.
Applying Pythagorean theorem to triangles AEO and ODB gives $4AC^2+BC^2=9\times 23^2$ and $AC^2+4BC^2=9\times 11^2$
Adding these gives $5(AC^2+BC^2)=9\times 650$
Replacing $AC^2+BC^2$ with $AB^2$ and rearranging gives $\frac{AB^2}9=130$
Also note that $OC=\frac{AB}3$, therefore:
$$OC^2=\frac{AB^2}9=130$$
