This is another tricky problem. The thing I found hard was to know how to make use of the fact that M is the midpoint of BC. Eventually, it occurred to me that this means that the triangles BAM, CAM have the same area (equal bases, same vertical height).

Denote by the three angles of the triangle. Let and let .

From B, draw a line BX to meet AK at X, such that . Then the triangles ABC, EBX are similar (because and ), and so .

Also, and hence the points A, B, C, X are concyclic (having equal angles in the same segment). Therefore (opposite angles of a cyclic quadrilateral are supplementary), and so . Also, and (angles in the same segment).

The area of triangle BAM is and the area of triangle CAM is . Since these are equal, it follows that .

Now look at the (vertical) distance from X to BC. This is equal to (since ). It is also equal to (since ). Therefore .

Putting together all the above facts, it follows by the RAR rule for similarity that the triangles EBX, EXC are similar. In particular, , as required.