The MRK triangle is isosceles, with the MR base. Angle K = 72, angle M = 54. a straight line parallel to the sides MP

The MRK triangle is isosceles, with the MR base. Angle K = 72, angle M = 54. a straight line parallel to the sides MP intersects the side of the RK at point A and the side of the MK at point B. Find the angle of the triangle ABK.

In general, the problem can be solved in one action, if you know the theory. With the correct construction of the drawing, you will get that the BMP angle corresponding to the KVA angle with parallel lines VA and MP with a secant MK. We know that for parallel lines and secant, the corresponding angles are equal. From this we conclude: the BMP angle = 54 degrees, which means that the angle of the KVA triangle will also be 54 degrees. If you want to find all the angles in a given triangle, then it is completely incomprehensible why the angle K is given, equal to 72 degrees. We can find the remaining KAV corner without it. If the CMR triangle is isosceles, then its base angles are equal. That is, the AWP angle is also 54 degrees. It is, again, corresponding to the KAV angle. Just at a different secant (CR). Accordingly, the remaining KAV angle in the KBA triangle = 54 degrees.
If you still use the angle K of 72 degrees, then you can take the axiom that the sum of the angles in any triangle = 180 degrees. Then subtract these 72 from 180, and then 54 – the degree measure of the KVA angle. So again we get 54 degrees – the angle KAV.
As I understand it, the problem can be solved in as many as 4 ways.