, 2011

Let $D'$ be the midpoint of the segment $AB$, and let $M$ be the midpoint of $BC$. By symmetry at line $AM$, the point $D'$ has to lie on the circle $BCD$. Since the $\operatorname{arcs} D'E$ and $ED$ of that circle are equal, we have $\angle ABI = \angle D'BE = \angle EBD = IBK$, so $I$ lies on the angle bisector of $\angle ABK$. For this reason it suffices to prove in the sequel that the ray $AI$ bisects the angle $\angle BAK$.

From
$$\angle DFA = 180^\circ - \angle BFA = 180^\circ - \angle BEA = \angle MEB = \frac{1}{2} \angle CEB = \frac{1}{2} \angle CDB$$
we derive $\angle DFA = \angle DAF$ so the triangle $AFD$ is isosceles with $AD = DF$.

Applying Menelaus's theorem to the triangle $ADF$ with respect to the line $CIK$, and applying the angle bisector theorem to the triangle $ABF$, we infer
$$1 = \frac{AC}{CD} \cdot \frac{DK}{KF} \cdot \frac{FI}{IA} = 2 \cdot \frac{DK}{KF} \cdot \frac{BF}{AB} = 2 \cdot \frac{DK}{KF} \cdot \frac{BF}{2 \cdot AD} = \frac{DK}{KF} \cdot \frac{BF}{AD}$$
and therefore
$$\frac{BD}{AD} = \frac{BF + FD}{AD} = \frac{BF}{AD} + 1 = \frac{KF}{DK} + 1 = \frac{DF}{DK} = \frac{AD}{DK} .$$
It follows that the triangles $ADK$ and $BDA$ are similar, hence $\angle DAK = \angle ABD$. Then
$$\angle IAB = \angle AFD - \angle ABD = \angle DAF - \angle DAK = \angle KAI$$
shows that the point $K$ is indeed lying on the angle bisector of $\angle BAK$.

It can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\angle ABK$. Here we restrict ourselves to proving that $KI$ bisects $\angle AKB$.

Denote the circumcircle of triangle $BCD$ and its center by $\omega_1$ and by $O_1$, respectively. Since the quadrilateral $ABFE$ is cyclic, we have $\angle DFE = \angle BAE = \angle DAE$. By the same reason, we have $\angle EAF = \angle EBF = \angle ABE = \angle AFE$. Therefore $\angle DAF = \angle DFA$, and hence $DF = DA = DC$. So triangle $AFC$ is inscribed in a circle $\omega_2$ with center $D$.

Denote the circumcircle of triangle $ABD$ by $\omega_3$, and let its center be $O_3$. Since the $\operatorname{arcs} BE$ and $EC$ of circle $\omega_1$ are equal, and the triangles $ADE$ and $FDE$ are congruent, we have $\angle AO_1B = 2 \angle BDE = \angle BDA$, so $O_1$ lies on $\omega_3$. Hence $\angle O_3O_1D = \angle O_3DO_1$.

The line $BD$ is the radical axis of $\omega_1$ and $\omega_3$. Point $C$ belongs to the radical axis of $\omega_1$ and $\omega_2$, and $I$ also belongs to it since $AI \cdot IF = BI \cdot IE$. Hence $K = BD \cap CI$ is the radical center of $\omega_1$, $\omega_2$, and $\omega_3$, and $AK$ is the radical axis of $\omega_2$ and $\omega_3$. Now, the radical axes $AK$, $BK$ and $IK$ are perpendicular to the central lines $O_3D$, $O_3O_1$ and $O_1D$, respectively. By $\angle O_3O_1D = \angle O_3DO_1$, we get that $KI$ is the angle bisector of $\angle AKB$.

Again, let $M$ be the midpoint of $BC$. As in the previous solutions, we can deduce $\angle ABI = \angle IBK$. We show that the point $I$ lies on the angle bisector of $\angle KAB$.

Let $G$ be the intersection point of the circles $AFC$ and $BCD$, different from $C$. The lines $CG$, $AF$, and $BE$ are the radical axes of the three circles $AGFC$, $CDB$, and $ABFE$, so $I = AF \cap BE$ is the radical center of the three circles and $CG$ also passes through $I$.

The angle between line $DE$ and the tangent to the circle $BCD$ at $E$ is equal to $\angle EBD = \angle EAF = \angle ABE = \angle AFE$. As the tangent at $E$ is perpendicular to $AM$, the line $DE$ is perpendicular to $AF$. The triangle $AFE$ is isosceles, so $DE$ is the perpendicular bisector of $AF$ and thus $AD = DF$. Hence, the point $D$ is the center of the circle $AFC$, and this circle passes through $M$ as well since $\angle AMC = 90^\circ$.

Let $B'$ be the reflection of $B$ in the point $D$, so $ABCB'$ is a parallelogram. Since $DC = DG$ we have $\angle GCD = \angle DBC = \angle KB'A$. Hence, the quadrilateral $AKCB'$ is cyclic and thus $\angle CAK = \angle CB'K = \angle ABD = 2 \angle MAI$. Then
$$\angle IAB = \angle MAB - \angle MAI = \frac{1}{2} \angle CAB - \frac{1}{2} \angle CAK = \frac{1}{2} \angle KAB$$
and therefore $AI$ is the angle bisector of $\angle KAB$.