Sometimes its Harder the other way Part 2

I spent some time thinking about what initially looked like a very simple triangle congruence problem last night which I’ve outlined below. Given a perpendicular angle bisector of a triangle, show that it intersects the bottom at its median. Turn the problem on its head, (Given a perpendicular bisector of a triangle that intersects the median prove its perpendicular) and it suddenly become quite a bit trickier since you lose the obvious way to show the triangles are congruent.

What I like here is that its very similar structurally to this problem: sometimes-one-direction-is-a-lot-tricker-than-the-other

Both problems have the same asymmetric complexity. Both are solvable by essentially reflecting above or below the original figure.  Even better, this version is not as hard the previous one which would make  for a very nice progression if presented in sequence.

(Apology: I tried doing the whole explanation in Geogebra which doesn’t scale as well as I intended. I recommend zooming in to read the figures)