Bulgarian National Mathematical Olympiad

Let $I_e,I_f,J_e,J_f$ be the incenters of $△ABE,△ABF,△CDE,△CDF$, respectively. Let $P=A{I}_e\cap B{I}_f$, $Q=A{I}_f\cap B{I}_e$, $U=C{J}_e\cap D{J}_f$, and $V=C{J}_f\cap D{J}_e$ be the excenter of $△ABC$ opposite to $A$, the excenter of $△ABD$ opposite to $B$, the incenter of $△ACD$, and the incenter of $△BCD$, 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 $△TAQ\sim △TUD$; therefore, $△TAU\sim △TQD$ and $\angle ATQ=\angle (AU,QD)=\frac{1}{2}\widehat{AD}=\angle A{I}_{e}Q$. It follows that $T$ lies on the circumcircle of $△A{I}_{e}Q$. Analogously, $T$ lies on the circumcircle of $△B{I}_{e}P,△C{J}_{e}V$, and $△D{J}_{e}U$.

We have, then, $\angle ATB=\angle AT{I}_e+\angle I_{e}TB=\angle AQB+\angle APB=\frac{1}{2}\widehat{AB}+\frac{1}{2}\widehat{AB}=\widehat{AB}$, showing that $T$ lies on $k$.



Let $T^{\prime }\in I_f{I}_e$ so that $I_{f}A\cdot I_{f}Q=I_f{I}_e\cdot I_f{T}^{\prime }=I_{f}B\cdot I_{f}P$. Then $T^{\prime }$ is a point other than $I_e$ which lies on the circumcircles of both $△A{I}_{e}Q$ and $△B{I}_{e}P$; therefore, $T^{\prime }\equiv T$ and $T\in I_e{I}_f$.

Analogously, $T\in J_e{J}_f$, and we are done.