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 2017-09-19 09:56:07