### What Might This Be About?

26 January 2016, Created with GeoGebra

### Problem

In $\Delta ABC\;$ $MN\;$ is an antiparallel, $M\in AC,\;$ $N\in BC;\;$ $O\;$ the circumcenter. Then $MN\perp OC.$

### Proof

Let $B'\;$ be the midpoint of $AC,;$ $D$ the intersection of $MN\;$ and $OC.\;$ Then triangles $CDM\;$ and $CB'O\;$ are similar. Indeed, they share an angle at $C.\;$ Also, since $AOC$ is a central angle in $(ABC),\;$ $\angle AOC=2\angle ABC,\;$ implying $\angle B'OC=\angle ABC=\angle CDM,\;$ the latter because $MN\;$ is an antiparallel.

So the triangles are similar. $\angle CB'O=90^{\circ},\;$ hence, $\angle CDM=90^{\circ}.$

### Remarks

Grégoire Nicollier has observed that the above result is a direct consequence of the fact that circumcenter and orthocenter are isogonal conjugates: when you reflect $\Delta ABC\;$ in the angle bisector of $C,\;$ the altitude through $C\;$ is mapped to the circumdiameter through $C.\;$ In addition, he noted that in a $\Delta ABC,\;$ the bisector of angle $A\;$ and the perpendicular bisector of side $BC\;$ meet on the circumcircle (an immediate consequence of the inscribed angle theorem.)

• 