Bulgarian National Mathematical Olympiad

Let $I_e, I_f, J_e, J_f$ be the incenters of $\triangle ABE, \triangle ABF, \triangle CDE, \triangle CDF$, respectively. Let $P = AI_e \cap BI_f$, $Q = AI_f \cap BI_e$, $U = CJ_e \cap DJ_f$, and $V = CJ_f \cap DJ_e$ be the excenter of $\triangle ABC$ opposite to $A$, the excenter of $\triangle ABD$ opposite to $B$, the incenter of $\triangle ACD$, and the incenter of $\triangle BCD$, respectively.

We have $\angle APB = \frac{1}{2} \widehat{AB} = \angle AQB$; therefore, $ABPQ$ is cyclic. Analogously, $CDUV$ is cyclic.

We have $\angle APB = \frac{1}{2} \widehat{AB} = \angle UCV$, $\angle AQB = \frac{1}{2} \widehat{AB} = \angle UDV$, $\angle APQ = \angle ABQ = \frac{1}{2} \widehat{AD} = \angle UCD$, and $\angle BQP = \angle BAP = \frac{1}{2} \widehat{BC} = \angle VDC$. Therefore, the figures $ABPQ$ and $UVCD$ are similar and identically oriented. Let $T$ be their center of similitude. (I.e., let $T$ be the fixed point of the unique similitude which maps $A, B, P, Q$ onto $U, V, C, D$, respectively.)

We have $\triangle TAQ \sim \triangle TUD$; therefore, $\triangle TAU \sim \triangle TQD$ and $\angle ATQ = \angle (AU, QD) = \frac{1}{2} \widehat{AD} = \angle AI_eQ$. It follows that $T$ lies on the circumcircle of $\triangle AI_eQ$. Analogously, $T$ lies on the circumcircle of $\triangle BI_eP, \triangle CJ_eV$, and $\triangle DJ_eU$.

We have, then, $\angle ATB = \angle ATI_e + \angle I_eTB = \angle AQB + \angle APB = \frac{1}{2} \widehat{AB} + \frac{1}{2} \widehat{AB} = \widehat{AB}$, showing that $T$ lies on $k$.

Let $T' \in I_f I_e$ so that $I_f A \cdot I_f Q = I_f I_e \cdot I_f T' = I_f B \cdot I_f P$. Then $T'$ is a point other than $I_e$ which lies on the circumcircles of both $\triangle AI_eQ$ and $\triangle BI_eP$; therefore, $T' \equiv T$ and $T \in I_e I_f$.

Analogously, $T \in J_e J_f$, and we are done.