Let ABC be an acute triangle with altitudes AD and BE. The
intersection of AD and BE is H. Points F and G make CADF
and CBEG into parallelograms. M is the midpoint of FG.
Ray MC intersects the circumcircle of ΔABC again at point N.

Prove that ANBH is a parallelogram.
  

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