Stewart's Theorem and Medians In triangle $ABC$, the median from vertex $A$ and the median from vertex $B$ intersect at right angles to each other at point $P$. If $AB = 12$, $BC = 13$, and $AC = 11$, what is the length of $AP$?
My attempt at this problem utilized Stewart's Theorem, where the length of the median BC is $19/2$ . Since the medians intersect perpendicularly at P, they cut each other into $2:1$ ratios. The length of is thus $(19/2)(2/3) = 19/3$, but this was marked incorrect. Am I missing something? Any help would be appreciated.

In a triangle with sides $a$, $b$, and $c$, the medians to the sides of length $a$ and $b$ are perpendicular if and only if $a^2+b^2=5c^2$. But $11^2+13^2\ne5\cdot12^2$, so something is wrong with the question. https://en.wikipedia.org/wiki/Median_(geometry)#Other_properties
