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International Mathematical Olympiad Spain

Geometry Difficulty 8.6 Shortlist Prove it IMO

Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$ , assume that there exist points $E$ on line $B C$ outside segment $B C$ , and $F$ inside segment $A D$ , such that $\angle D A E=\angle C B F$ . Denote by $I$ the point of intersection of $C D$ and $E F$ , and by $J$ the point of intersection of $A B$ and $E F$ . Let $K$ be the midpoint of segment $E F$ ; assume it does not lie on line $A B$ .
Prove that $I$ belongs to the circumcircle of $A B K$ if and only if $K$ belongs to the circumcircle of $C D J$ .

Solution

Assume that the disposition of points is as in the diagram.
Since $\angle E B F=180^{\circ}-\angle C B F=180^{\circ}-\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \cdot J B=F J \cdot J E$ . In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $I J \cdot J K=F J \cdot J E$ . Expressing $I J=I F+F J$ , $J E=F E-F J$ , and $J K=\frac{1}{2} F E-F J$ , we find that $I$ belongs to the circumcircle of $A B K$ if and only if
$F J=\frac{I F \cdot F E}{2 I F+F E}$
Since $A E B F$ is cyclic and $A B, C D$ are parallel, $\angle F E C=\angle F A B=180^{\circ}-\angle C D F$ . Then $C D F E$ is also cyclic, yielding $I D \cdot I C=I F \cdot I E$ . It follows that $K$ belongs to the circumcircle of $C D J$ if and only if $I J \cdot I K=I F \cdot I E$ . Expressing $I J=I F+F J$ , $I K=I F+\frac{1}{2} F E$ , and $I E=I F+F E$ , we find that $K$ is on the circumcircle of $C D J$ if and only if
$F J=\frac{I F \cdot F E}{2 I F+F E} .$
The conclusion follows.

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