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Team Selection Test for IMO, 2023

Geometry Difficulty 8.4 Shortlist Prove it Turkey

In the interior of a trapezoid $ABCD$ with $AB \parallel CD$ a point $T$ is chosen such that $\angle ATD = \angle CTB$ . The second intersection point of the line $AT$ with the circumcircle of $ACD$ is $K$ . The second intersection point of the line $BT$ with the circumcircle of $BCD$ is $L$ . Show that $KL \parallel AB$ .

Solution

Let the lines $AT$ and $BT$ intersect the line $CD$ at the points $A_1$ and $B_1$ , and intersect the circle $(TCD)$ at the points $A_2$ and $B_2$ (beside $T$ ). Now one has $\overline{A_2D} = \overline{CB_2}$ on the circle $(TCD)$ , which implies $A_2B_2 \parallel CD$ . On the other hand $A_1T \cdot A_1A_2 = A_1C \cdot A_1D = A_1A \cdot A_1K$ , hence one gets $A_1K/A_1A_2 = A_1T/A_1A$ and one analogously gets $B_1L/B_1B_2 = B_1T/B_1B$ . Now $A_1T/A_1A = B_1T/B_1B$ , therefore $A_1K/A_1A_2 = B_1L/B_1B_2$ , which implies $KL \parallel A_2B_2$ .

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