Estonian Mathematical Olympiad

We have $\angle KBE=90^{\circ }$ as $BE$ and $BK$ are the internal and external angle bisectors at the same vertex (Fig. 43). We will now show $\angle BKC=90^{\circ }$. Let $X$ be a point on line $AB$ such that $CX\parallel BE$; then the external angle bisector at vertex $B$ of triangle $ABC$ is also perpendicular to line $CX$ (Fig. 44). Denote $\angle ABC=\beta$. Then $\angle BCX=\angle CBE=\frac{\beta }{2}$ and $\angle XBC=180^{\circ }-\angle ABC=180^{\circ }-\beta$. Therefore, $\angle BXC=180^{\circ }-(\angle BCX+\angle XBC)=\frac{\beta }{2}$, so the triangle $XBC$ is isosceles with apex at $B$. Thus, the altitude from vertex $B$ of triangle $XBC$ bisects the base $CX$. As previously established, this altitude lies on the external angle bisector at vertex $B$ of triangle $ABC$. Let $k=\frac{|FX|}{|FB|}=\frac{|FC|}{|FI|}$. A homothety with center $F$ and ratio $k$ sends points $B$ and $I$ to points $X$ and $C$, respectively, so it sends the midpoint $M$ of segment $BI$ to the midpoint of segment $CX$. Therefore, the line $FM$ passes through the midpoint of segment $CX$. Consequently, the intersection point $K$ of line $FM$ and the external angle bisector at vertex $B$ of triangle $ABC$ lies at the midpoint of segment $CX$, and $\angle BKC=90^{\circ }$, as we wanted to show. Similarly, we can show that $\angle BLC=90^{\circ }$. Therefore, points $K$ and $L$ lie on the circle with diameter $BC$. The statement of the problem follows.