# Symmedian via Parallel Transversal and Two Circles

### Problem

Let $MN\;$ be a transversal parallel to the side $BC\;$ of $\Delta ABC,\;$ with $M\;$ on $AB\;$ and $N\;$ $AC.\;$ The lines $BN\;$ and $CM\;$ meet at point $P.\;$ The circumcircles of triangles $BMP\;$ and $CNP\;$ meet at two distinct points $P\;$ and $Q.\;$

Prove that the line $AQ\;$ is the $A\text{-symmedian}\;$ of $\Delta ABC.$

### Proof

Consider triangles $BQM\;$ and $NQC:$

(1)

$\angle BQM=\angle BPM=\angle CPN=\angle CQN.$

Also,

(2)

$\angle MBQ=180^{\circ}-\angle MPQ=\angle CPQ=\angle CNQ.$

Form (1) and (2), triangles $BQM\;$ and $NQC\;$ are similar and their respective elements are proportional. It follows that

$\displaystyle\frac{dist(Q,AB)}{dist(Q,AC)}=\frac{dist(Q,BM)}{dist(Q,CN)}=\frac{BM}{CN}=\frac{AB}{AC},$

implying that $AQ\;$ is indeed the symmedian in $\Delta ABC\;$ through vertex $A.$

### References

1. Sammy Luo and Cosmin Pohoata, Let's Talk About Symmedians!, Mathematical Reflections 4 (2013), 1-11