Locus of all points C Let AB be a given line segment. Find all possible points C in the plane such that in the
triangle ABC, the altitude from A and the median from B have the same length.
I let AD and BE be an altitude and median respectively. I then drew an altitude from E to BC to get that CBE must equal 30. I don't know how to finish after this step though.

Let $A'$ be such that $B$ is the mid point of the segment $AA'$. (So reflect $A$ w.r.t. $B$ to get $A'$.) It is a fixed point, given the fixed points $A,B$.
Then with the notations from the OP $BE$ is mid line in $\Delta ACA'$, so $BE\|A'C$, so
$$
\widehat{BCA'}=30^\circ\ .
$$
This restricts $C$ to live on the corresponding two arcs with the chord $BA'$ (passing through the one or the other point $C^*$ that makes $\Delta ABC^*$ equilateral). In a picture:

Reciprocally, for each point $C$ on the one or the other arc based on the chord $BA'$, we consider $E$ the mid point of the segment $AC$, then $\widehat{CBE}$ is $30^\circ$, and the same argument from the OP insures that $BE$ is equal to the altitude from $A$ in $\Delta ABC$.
