NMO Selection Tests

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Romania

Let $ABCD$ be a convex quadrilateral with the property that the circles having the segments $AB$ and $CD$ as diameters are tangent externally at a point $M$ , different from the intersection point of the diagonals of the quadrilateral.
Let $K$ be the second point of intersection of the circumcircle of triangle $AMC$ with the line determined by $M$ and the midpoint of segment $AB$ , and let $L$ be the second point of intersection of the circumcircle of triangle $BMD$ with the line determined by $M$ and the midpoint of segment $CD$ .
Prove that $|MK - ML| = |AB - CD|$ .

Solution

Let $F$ and $E$ be the midpoints of chords $ML$ and $MK$ , respectively. Then $O_3F \perp ML$ and $O_4E \perp MK$ . Since $O_4O_1O_2O_3$ is cyclic with $\angle O_4O_2O_3 = \angle O_4O_1O_3 = 90^\circ$ , the midpoint $X$ of segment $O_3O_4$ is the center of the circumscribed circle of $O_4O_1O_2O_3$ . Moreover, if $T$ denotes the foot of the perpendicular from $X$ onto $O_1O_2$ , then $TO_1 = TO_2$ .

In the right trapezoid $O_4EFO_3$ , since $X$ is the midpoint of $O_3O_4$ and $XT \perp EF$ , it follows that $XT \parallel EO_4 \parallel FO_3$ , so $XT$ is a midline and therefore $ET = TF$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.