Estonian Mathematical Olympiad

Let $O$ be the center of $\omega$ and $Q$ be the intersection point of lines $DO$ and $AT$. Since $ABCT$ is a trapezoid and a cyclic quadrilateral, it is an isosceles trapezoid and the points $A$ and $T$ are symmetric with respect Fig. 47 Fig. 48

to line $DO$ (Fig. 47). We have $\angle DKT=\angle AKT=\frac{1}{2}\angle AOT=\angle QOT=180^{\circ }-\angle DOT$, whereas $K$ and $O$ are on different sides of line $DT$, so point $O$ is on the circumcircle of triangle $KDT$.

Since $O$ is the point of intersection of the perpendicular bisectors of triangle $ABC$, we have $\angle OEA=\angle OFA=90^{\circ }$ (Fig. 48). Thus points $E$ and $F$ are on the circle with diameter $OA$ and $O$ is on the circumcircle of triangle $AEF$. Since triangle $ABC$ is acute, triangle $AEF$ is also acute (because $EF\parallel BC$), hence its center is inside the triangle. So point $O$ is on the shorter arc $EF$. Thus the circumcircles of triangles $KDT$ and $AEF$ intersect at $O$ on the shorter arc $EF$, so $O=L$.

Since $AL$ is a diameter of the circumcircle of triangle $AEF$ and point $M$ is on the circle, $ML$ and $AK$ are perpendicular (Fig. 49). Therefore $KL$ is a diameter of the circumcircle of triangle $KLM$. Since $KL$ is also a radius of circle $\omega$, the line through $K$ perpendicular to line $KL$ is tangent to both circles. Consequently circle $\omega$ and the circumcircle of triangle $KLM$ are tangent to each other.

Fig. 49