Can this be solved without trigonometry?
$AB$ is the base of an isosceles $\triangle ABC$. Vertex angle $C$ is $50^\circ$. Find the angle between the altitude and the median drawn from vertex $A$ to the opposite side.
I think I know how to do this using the Law of Cosines.
Call the side of the isosceles triangle $x$, $CA=CB=x$, and let $E$ be the midpoint of $BC$. Then in the triangle formed by the median, $\triangle CAE$, using the Law of Cosines:
$$AE = \sqrt{x^2 + \frac{x^2}4-2\frac{x^2}{2}\cos50^\circ}$$
From this you can find $AE$ in terms of $x$.
Then apply the Law of Cosines again to find $\angle CAE$ using $$CE=\frac{x}{2}= \sqrt{x^2 + AE^2-2xAE\cos\angle CAE}$$ and from $\cos\angle CAE$ you find $\angle CAE$, and the angle we want is $40^\circ-\angle CAE$.
But is it possible w/o trig?