$G$ is the centroid of $\triangle ABC$. $BE \cap CF = D$. Prove that $AD$ is the symmedian of $\triangle ABC$.

Question

$G$ is the centroid of $\triangle ABC$. $E$ and $F$ are points respectively on $(AGB)$ and $(AGC)$ such that $GE \parallel AB$ and $GF \parallel AC$ $(G \not\equiv E, G \not\equiv F)$. $BE \cap CF = D$. Prove that $AD$ is the symmedian of $\triangle ABC$.

We have that $AG$ is the median of $\triangle ABC$.

It needs to be sufficient to prove that $AG$ and $AD$ reflect one another over the internal bisector of angle $\angle CAB$, which means proving $$\angle BAG = \angle CAD \ (\angle CAG = \angle BAD)$$

Since $ABGE$ and $ACGF$ are concyclic parallelograms, we obtain that $$\left\{ \begin{align} \angle BAG = \angle ABE &\ (\angle BAE = \angle ABG)\\ \angle CAF = \angle ACG &\ (\angle CAG = \angle ACF) \end{align} \right.$$

We now need to demonstrate that $$\angle CAD = \angle ABE \ (\angle BAD = \angle ACF) \impliedby \triangle CDB \sim \triangle DCA \impliedby \left\{ \begin{align} \frac{CD}{DA} &= \frac{AD}{DB}\\ \angle CDA &= \angle ADB \end{align} \right.$$

It could also be observed that $AD \perp DO$ where $O$ is the circumcentre of $\triangle ABC$, although I am unsure about why.

Answer 1

Let $M$ be a point such that the quadrilateral $ABMC$ is harmonic. Let $N$ be the midpoint of $AM$.

By properties of harmonic quadrilaterals, $AM$ is symmedian of $ABC$ and $BC$ is symmedian of $MBA$. It follows that $$\angle NBA = \angle MBC = \angle MAC = \angle BAG = \angle EBA.$$ It follows that $N$ lies on $BE$. We can prove similarly that $N$ lies on $CF$. Therefore $N \equiv D$ and we are done because $AD$ coincides with $AM$ which is symmedian of $ABC$.