IMO Shortlist, 2007

Geometry Difficulty 8.1 Shortlist Prove it IMO

The diagonals of a trapezoid $ABCD$ intersect at point $P$ . Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\angle AQD = \angle CQB$ , and line $CD$ separates points $P$ and $Q$ . Prove that $\angle BQP = \angle DAQ$ .

Solution

Let $t = \frac{AD}{BC}$ . Consider the homothety $h$ with center $P$ and scale $-t$ . Triangles $PDA$ and $PBC$ are similar with ratio $t$ , hence $h(B) = D$ and $h(C) = A$ .

Let $Q' = h(Q)$ (see Figure 1). Then points $Q$ , $P$ and $Q'$ are obviously collinear. Points $Q$ and $P$ lie on the same side of $AD$ , as well as on the same side of $BC$ ; hence $Q'$ and $P$ are also on the same side of $h(BC) = AD$ , and therefore $Q$ and $Q'$ are on the same side of $AD$ . Moreover, points $Q$ and $C$ are on the same side of $BD$ , while $Q'$ and $A$ are on the opposite side (see Figure above).

By the homothety, $\angle AQ'D = \angle CQB = \angle AQD$ , hence quadrilateral $AQ'QD$ is cyclic. Then
$\angle DAQ = \angle DQ'Q = \angle DQ'P = \angle BQP$
(the latter equality is valid by the homothety again).

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