, 2021

Refer to figure 12.

As $O$ is the circumcenter of $\triangle ABC$ it follows that line segments $OB$ and $OC$ are concurrent. As $D$ is the reflection of $O$ across $BC$ it follows that segments $OB$ and $DB$ are concurrent and segments $OC$ and $DC$ are concurrent. It follows that $OBDC$ is a rhombus and therefore also a parallelogram. Similarly $OCEA$ is also a parallelogram.

As lines $BD$ and $OC$ are parallel and lines $OC$ and $AE$ are parallel it follows from the transitivity of parallelism that lines $BD$ and $AE$ are parallel.

As lines $OB$ and $CD$ are parallel and lines $OA$ and $CE$ are parallel it follows from Desargues's theorem (the Euclidean plane is a translation plane) that lines $AB$ and $ED$ are parallel. We have hence shown that $ABDE$ is a parallelogram. Similarly it can be shown that $AFDC$ is also a parallelogram.

Let $P$ be the intersection point of lines $AD$ and $BE$. As $ABDE$ is a parallelogram it follows that $P$ is the midpoint of segment $AD$. As $P$ is the midpoint of segment $AD$ and $AFDC$ is a parallelogram it follows that line $DF$ passes through $P$. We have therefore demonstrated that $P$ is a common point on $AD$, $BE$ and $CF$, that is lines $AD$, $BE$ and $CF$ are concurrent. $\square$