Croatian Mathematical Olympiad

Instead of the original formulation, we will solve the following equivalent problem: Let $ABCD$ be an isosceles trapezium with bases $\overline{AB}$ and $\overline{CD}$, and let its diagonals meet at the point $S$, as in the original formulation. Additionally, let $K$ be a point on the line $AD$ such that $SK\parallel AB$, and let $M$ be the other intersection of $AD$ with the circle circumscribed to the triangle $BCK$. We need to prove that $M$ is the midpoint of the leg $\overline{AD}$.



Let $M^{\prime }$ be a point on the line $BC$ such that $M{M}^{\prime }\parallel AB$, and let $T$ be the intersection of the lines $CK$ and $AB$. Since the trapezium $ABCD$ is isosceles and the quadrilateral $BCKM$ is cyclic, we have $\angle DA{M}^{\prime }=\angle CBM=\angle MKT$, and therefore $A{M}^{\prime }\parallel TC$. Using similarity of triangles, we obtain $|AT|:|CD|=|KA|:|KD|=d(S,AB):d(S,CD)=|AB|:|CD|$. Hence $|AT|=|AB|$ and $|M^{\prime }B|=|M^{\prime }C|$. Thus the point $M^{\prime }$ is the midpoint of the leg $\overline{BC}$ and, of course, $M$ is the midpoint of the leg $\overline{AD}$.