Problem $(a)$ Solution: Image not found The quadrilateral $AIGC$ is cyclic when the points $A,I,G$ and $C$ are concyclic. Hence, to prove the statement we just need to prove that angles $GAI$ and $ICG$ are right angled with diameter $GI$ of course. - To prove that angles $GAI$ and $ICG$ are right angled, it's sufficient to prove that $\angle GAF=\left( 90^\circ - \frac{\alpha}{2}\right) $ and $\angle GCF=\left(90^\circ - \frac{\gamma}{2}\right)$, respectively. $\bullet$ It's obviously seen that $\triangle AFD$ and $\triangle CFE$ are isosceles. According to the condition of the problem we have $\overline{AG} \parallel \overline{DF}$ which implies \[\angle GAF= \angle AFD=\angle FDA \ \ \ \cdots (1) \] For same reasoning; parallelism of $\overline{GC}$ and $\overline{FE}$ we have: \[\angle GCF= \angle CFE=\angle FEC \ \ \ \cdots (2) \] Therefore, in $\triangle AFD$ we have: $2\cdot \angle AFD + \alpha = 180^\circ \implies \angle AFD = \left( 90^\circ - \frac{\alpha}{2}\right)$ And since $\angle AFD=\angle GAF$, see from $(1)$ we obtain: $\angle GAF = \left( 90^\circ - \frac{\alpha}{2}\right)$. Therefore, $\angle GAI = 90^\circ$ Similarly, in triangle $CFE$ we have: $ 2\cdot \angle CFE + \gamma = 180^\circ \implies \angle CFE = \left( 90^\circ - \frac{\gamma}{2}\right)$ And since $\angle CFE=\angle GCF$, see from $(2)$, we obtain: $\angle GCF=\left(90^\circ - \frac{\gamma}{2}\right)$. Therefore, $\angle ICG = 90^\circ$ $\bullet$ Hence, angles $GAI$ and $ICG$ are right angled, thus the points $A,I,G$ and $C$ are concyclic which implies that the quadrilateral $AIGC$ is cyclic.

WakeUp wrote: $(b)$ Prove that the points $B,I,G$ are collinear. Problem $(b)$. Solution If $G,I,B$ are collinear points, then angle between $\overline{IG}$ and $\overline{IB}$ must be equal to $180^\circ$. From the case in $(a)$ from triangle $GAI$ we have: $\angle AIG = \left(90^\circ - \frac{\gamma}{2}\right)$, because $\frac{\gamma}{2}=\angle AGI$ as angles from the same segment in circle. Similarly, from triangle $AIB$ we have the relation \[\angle AIB=180^\circ - \left(\frac{\alpha+\beta}{2}\right) = 180^\circ - \left(90^\circ -\frac{\gamma}{2}\right) = 90^\circ + \frac{\gamma}{2}\] Therefore angle between $\overline{IG}$ and $\overline{IB}$ is equal to $\angle AIG+\angle AIB = 180^\circ$, hence $B,I,G$ are collinear.

Dear Mathlinkers, this situation has been studied by Mention... G, usually noted Ib, is the B-excenter of ABC ; hence b) G, A, I and C is on the Mention's B-circle of ABC ; its center is on the circumcircle of ABC and more precesely the midpoint of the arch AC which doesn't contain B. Sincerely Jean-Louis

a)$ADIF$ is cyclic with diameter $AI$,but $\angle FAG=\angle AFD$,also $\angle AFD=\angle ADF$,so $\angle FAG=\angle ADF$, hence $AG$ is tangent to the circumcircle of $ADIF$ at $A$,so $\angle IAG=90$,similarly $\angle ICG=90$,hence $AICG$ is cyclic. b)$\angle BIC=90+\angle A/2$,$\angle CIG=\angle CAG=\angle FDA=90-\angle A/2$,so $\angle BIC+\angle CIG=180$.