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IMO Team Selection Test

Geometry Difficulty 7.7 National Olympiad, round 2 Prove it North Macedonia

Let $ABCD$ be an inscribed quadrangle, and let $BC$ and $AD$ intersect at point $P$ . The point $Q$ belongs to the line $BP$ in such a way that $\overline{PQ} = \overline{BP}$ , and $CAQR$ and $DBCS$ are parallelograms. Prove that the points $C, Q, R$ and $S$ are concyclic.

Solution

Obviously, it is enough to show that
$\angle RQC = \angle RSC$
(*)
From the conditions of the problem we have
$\angle RQC = \angle ACQ = \angle ACB = \angle ADB$
(1)
We choose a point $T$ , such that $QABT$ is a parallelogram. Then $\overline{BT} = \overline{AQ} = \overline{CR}$ and $\overline{BD} = \overline{CS}$ . According to that, $\Delta BTD \cong \Delta CRS$ from where we get
$\angle RSC = \angle TDB$
(2)
On the other hand, the point $P$ is the midpoint of $BQ$ in the parallelogram $ABTQ$ and therefore is the midpoint of segment $AT$ . Now, $\angle TDB = \angle ADB$ , so from (1) and (2) we get (*).

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