66th Belarusian Mathematical Olympiad

Let I be the incenter of the triangle ABC. Let $\Gamma$ denote the circle passing through $A$, $K$, $L$. We have $$\begin{array}{rl}\angle AKD=\angle EKD& =180^{\circ }-\angle DME=\angle DMF=180^{\circ }-\angle FLD=\\ & =180^{\circ }-\angle ALD.\end{array}$$ Therefore, $D$ lies on $\Gamma$. So, to prove that $\Gamma$ touches $BC$ it suffices to show that $\angle DAL=\angle BDL$.



Since $AE=AF$ and $EM=MF$, it follows that $AM\perp EF$ and $AM$ is the bisector of the angle $EAF$, i.e., the points $A$, $M$, $I$ lie on the same line. We have $$\begin{array}{rl}\angle BDL& =90^{\circ }-\angle IDL=90^{\circ }-(\angle IDM+\angle MDL)=\\ & =(90^{\circ }-\angle MDL)-\angle IDM=[\angle MDL=180^{\circ }-\angle MFL=\angle MFA]=\\ & =(90^{\circ }-\angle MFA)-\angle IDM=\angle FAM-\angle IDM=\\ & =\angle DAL+\angle DAM-\angle IDM.\end{array}\text{\hspace{1em}(1)}$$

Since the triangle $AFI$ is the right-angled triangle and $FM\perp AI$, we have $I{F}^2=IM\cdot IA$. Since $IF=ID$, we obtain $I{D}^2=IM\cdot IA$. By the Power of a Point Theorem, the circle passing through the points $A$, $M$, $D$ touches the line $ID$ at $D$, then $\angle IDM=\angle DAM$. From (1) it follows that $\angle DAL=\angle BDL$ as required.