Angle between the median and altitude to one side of an isosceles triangle 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?

As @quasi's answer suggests, the target angle almost-certainly isn't rational, so avoiding trig is unlikely.
That said, there's a pretty quick trigonometric approach to the target:

Let $s$ be the triangle's half-leg, $M$ the midpoint of $\overline{BC}$, and $N$ the foot of the altitude from $A$. Then, in right $\triangle ACN$ we have
$$|AN|=2s\sin C \qquad |NC|=2s\cos C$$
Thus,

$$\tan\theta =\frac{|MN|}{|AN|}=\frac{|CN|-|CM|}{|AN|} = \frac{2s\cos C-s}{2s\sin C} = \frac{2\cos C-1}{2\sin C} $$

For $C=50^\circ$, this gives $\theta=10.558\ldots^\circ$.
