Saudi Arabian Mathematical Competitions, 2015

First solution. Let $P$ be the intersection point of the bisector of the angle $\angle BAC$ with the segment $EF$. Because both $I$ and $P$ are on the bisector of $\angle BAC$, the problem is equivalent to showing that $AP = AI$.
Because $DE$ is the bisector of $\angle ADB$ we have from the bisector theorem
$$\frac{AE}{AD} = \frac{BE}{BD} = \frac{AB}{AD + BD}$$
Because $DF$ is the bisector of $\angle CDA$ we have from the bisector theorem
$$\frac{AF}{AD} = \frac{CF}{CD} = \frac{AC}{AD + CD}.$$
We deduce that
$$\begin{aligned} \frac{1}{AE} + \frac{1}{AF} & = \frac{AD \cdot AC + BD \cdot AC + AD \cdot AB + CD \cdot AB}{AB \cdot AC \cdot AD} \\ & = \frac{AD \cdot AC + AD \cdot AB + AD \cdot BC}{AB \cdot AC \cdot AD} = \frac{AB + BC + CA}{AB \cdot AC} \end{aligned}$$
from Ptolemy's theorem applied to the cyclic quadrilateral $ABDC$.
Now from the areas of triangles $AEP$, $APF$ and $AEF$, we have
$$\frac{1}{2} AE \cdot AP \sin \frac{\alpha}{2} + \frac{1}{2} AF \cdot AP \sin \frac{\alpha}{2} = \frac{1}{2} AE \cdot AF \sin \alpha,$$
from which we deduce
$$\begin{gathered} AP = \frac{2 AE \cdot AF \cos \frac{\alpha}{2}}{AE + AF} = \frac{2 AB \cdot AC \cos \frac{\alpha}{2}}{AB + BC + CA} = \frac{AB \cdot AC \cos \frac{\alpha}{2}}{s} \\ = \frac{4R \Delta \cos \frac{\alpha}{2}}{s \cdot BC} = \frac{2r \cos \frac{\alpha}{2}}{\sin \alpha} = \frac{r}{\sin \frac{\alpha}{2}} = AI, \end{gathered}$$

Saudi Arabia Mathematical Competitions 2015
where $\alpha = \angle BAC$, $s = \frac{AB + BC + CA}{2}$ is the semi-perimeter of triangle $ABC$, $R$ its circumradius, $r$ its inradius and $\Delta = \frac{AB \cdot BC \cdot CA}{4R} = rs$ its area.

Second Solution. Let $B'$ and $B''$ be the intersection points of the bisector of the angle $\angle CBA$ with the side $BC$ and the $\text{arc BC}$. Denote $\beta = \angle CBA$, $\gamma = \angle ACB$ and $\theta = \angle BAD$.
We have $\angle B'B''F = \angle BB''D = \angle BAD = \theta$ and $\angle FB'B'' = \angle CAB'' + \angle AB''B = \angle CBB'' + \angle ACB = \frac{\beta}{2} + \gamma$. We deduce from sine law in triangle $B'FB''$ that
$$\frac{B'F}{FB''} = \frac{\sin \theta}{\sin \left(\frac{\beta}{2} + \gamma\right)}$$
We have $\angle AB''F = \angle AB''B + \angle BB''D = \angle ACB + \angle BAD = \gamma + \theta$ and $\angle FAB'' = \angle CAB'' = \angle CBB'' = \frac{\beta}{2}$. We deduce from sine law in triangle $AFB''$ that
$$\frac{FA}{FB''} = \frac{\sin (\gamma + \theta)}{\sin \frac{\beta}{2}}$$

We deduce that
$$\frac{\overline{B'F}}{\overline{FA}} = -\frac{\sin \theta \sin \frac{\beta}{2}}{\sin \left(\frac{\beta}{2} + \gamma\right) \sin (\gamma + \theta)}$$
Because $DE$ is the bisector of angle $\angle ADB$, we have from the theorem of bisectors
$$\frac{\overline{AE}}{\overline{EB}} = \frac{AD}{DB} = \frac{\sin \angle ABB''D}{\sin \angle DAB} = \frac{\sin (\gamma + \theta)}{\sin \theta}.$$
Because $AI$ is the bisector of angle $\angle BAB'$, we have from the theorem of bisectors and sine law in triangle $ABB'$
$$\frac{\overline{BI}}{\overline{IB'}} = \frac{AB}{AB'} = \frac{\sin \angle ABB'B}{\sin \angle B'BA} = \frac{\sin \left(\frac{\beta}{2} + \gamma\right)}{\sin \frac{\beta}{2}}.$$

Therefore
$$\frac{\overline{AE}}{\overline{EB}} \cdot \frac{\overline{BI}}{\overline{IB'}} \cdot \frac{\overline{B'F}}{\overline{FA}} = -1$$
and from the theorem of Menelaus applied to the triangle $ABB'$, the points $E$, $I$, $F$ are collinear.

Third Solution. Let $B'$ and $C'$ be the midpoints of the $\operatorname{arcs} \widehat{CA}$ and $\widehat{AB}$, respectively. Because $DB'$ and $DC'$ are the bisectors of the angles $\angle ADB$ and $\angle CDA$, respectively, point $E$ is the intersection of $AB$ and $DC'$, point $I$ is the intersection of $BB'$ and $C'C$, and point $F$ is the intersection of $B'D$ and $CA$. We deduce from Pascal's theorem applied to the cyclic hexagon $ABB'DC'C$ that the three points $E$, $I$ and $F$ are collinear.