Perpendicularity Associated to the Incenter

Let $ABC$ be any triangle and $I$ its Incenter. Call $E$ and $F$ the reflections of $I$ around the sides $AB$, $BC$, respectively. Now, denote $G$ the reflection of $I$ around the midpoint, $D$, of side $AB$. Prove that $AG\perp{EF}$.

Proof. This is just angle chasing. Indeed, call $L$ the intersection of $AG$ and $IE$, then, 

$$\angle{EPA}=180^\circ-\angle{FEI}-\angle{ELG}$$

Denote by $M$ and $N$ the intersection of the Incircle of $ABC$ with the sides $AB$, $BC$, respectively. As $BMIN$ is clearly a  right kite and $\triangle{EFI}$ is isosceles with $IE=IF=2r$, it follows

$$\angle{EPA}=180^\circ-\frac{180^\circ-\angle{MIN}}{2}-\angle{MLA}$$

$$\angle{EPA}=180^\circ-\left[\frac{180^\circ-(180^\circ-\angle{ABC})}{2}\right]-\angle{MLA}$$

$$\angle{EPA}=180^\circ-\frac{\angle{ABC}}{2}-\angle{MLA}$$

As $ID=DG$ and $AD=BD$ the quadrilateral $AGBI$ is a parallelogram, then, $\angle{IBA}=\angle{GAB}=\frac{\angle{ABC}}{2}$. Since $\angle{LMA}=90^\circ$, it follows that $\angle{MLA}=90^\circ-\frac{\angle{ABC}}{2}$. Finally, 

$$\angle{EPA}=180^\circ-\frac{\angle{ABC}}{2}-90^\circ+\frac{\angle{ABC}}{2}=90^\circ$$

$\square$