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.