Starting with a related problem instead: Let points F and G make ADCF and BGCE instead. The parallelograms are now rectangles  actually they are the same figures as in the problem but with the long sides bent against each other rather than stretched out. Also, point M is more usefully placed. Leave ray MC for the moment. The circumcircle is K with origin O.
Line AF intersects K at J. Line AD intersects K at P. JAP is a right angle on K
Line BE intersects K at S. Line BG intersects K at T. SBT is a right angle on K.
The same can be done for other points e.g. L on K and CF and Q on K and CG, the idea being that they are all diameters of K. Ray CM passes through the same point, namely O, and so CN is also a diameter.
We can now return to the problem as set, for if CN is a diameter then angles CAN and CBN must be right angles, enough to prove with the angles already known from construction that ANBH is a parallelogram.

Posted by broll
on 20170919 09:56:07 