Concurrence 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 $AB$. Let $O$ be the orthogonal projection of $C$ onto $AB$. Prove that $EF$, $BG$ and $CO$ are concurrent.

Proof. Supose that $EF$ and $CO$ meet at $S$. It suffices to show that $S$, $G$ and $B$ are collinear. It is easy to realize that $\angle{IEF}=\angle{IFE}=\frac{\angle{ABC}}{2}$. It follows 

$$\angle{CSF}=180^\circ-\angle{SFI}-\angle{IFC}-\frac{\angle{ACB}}{2}-\angle{ACB}-\angle{BAC}+90^\circ$$

$$\angle{CSF}=180^\circ-\frac{\angle{ABC}}{2}-\left(90^\circ-\frac{\angle{ACB}}{2}\right)-\frac{\angle{ACB}}{2}$$

$$-\angle{ACB}-\angle{BAC}+90^\circ=180^\circ-\frac{\angle{ABC}}{2}-\left(180^\circ-\angle{ABC}\right)=\frac{\angle{ABC}}{2}$$

Since $\angle{CBF}=\frac{\angle{ABC}}{2}$, the quadrilateral $CSBF$ is cyclic. Consequently, 

$$\angle{ABS}=180^\circ-\frac{\angle{ACB}}{2}-\left(\angle{BAC}-90^\circ\right)-\angle{ACB}-\frac{\angle{ABC}}{2}-\angle{ABC}=\frac{\angle{BAC}}{2}$$

Notice that $AIBG$ is a parallelogram, hence, $\angle{ABG}=\angle{BAI}=\frac{\angle{BAC}}{2}$. Since $\angle{ABG}=\angle{ABS}$, the points $S$, $G$ and $B$ are collinear and the proof is complete.