To show AXM and BXC are similar we need only show that they have two equal (congruent) angles.

1. Angles MXA and CXB are congriuent as they are vertical angles.

2. Angles AMX and CBX are congruent as they are alternate interior angles for the line BM between the parallel lines AD and BC.

Hence triangles AXM and BXC have two congruent angles (and so the third angles in each triangle are also congruent to one another as the angle sum in any triangle is 180 degrees) and so are similar.

RonL