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Croatia competition problems

Geometry Difficulty 6.2 National Olympiad Prove it Croatia

A trapezium $ABCD$ is given. The bisector of the leg $\overline{BC}$ intersects the leg $\overline{AD}$ at $M$ , while the bisector of $\overline{AD}$ intersects $\overline{BC}$ at $N$ .
Let $O_1$ and $O_2$ be the circumcentres of triangles $ABN$ and $CDM$ , respectively.
Prove that the line $O_1O_2$ bisects the segment $MN$ . (Stipe Vidak)

Solution

Denote by $P$ and $Q$ the midpoints of $\overline{BC}$ and $\overline{AD}$ , respectively.

We will show that the quadrilateral $ABNM$ is cyclic.
From $\angle MQN = \angle NPM = 90^\circ$ , we get that the quadrilateral $MNPQ$ is cyclic. This implies $\angle PQM + \angle MNP = 180^\circ$ .
The segment $\overline{QP}$ is the midsegment of the trapezium $ABCD$ , which means that it is parallel to $AB$ . From here we conclude that $\angle MAB = \angle DQP$ . Now we have
$\angle MAB = \angle DQP = 180^\circ - \angle PQM = \angle MNP = 180^\circ - \angle BNM,$
which is enough to conclude that the quadrilateral $ABNM$ is cyclic.
Analogously, we show that the quadrilateral $MNCD$ is cyclic.
This shows that the segment $\overline{MN}$ is simultaneously a chord for both circumscribed circles of triangles $ABN$ and $CDM$ . We conclude that the line $O_1O_2$ , connecting the centres of these circles, must bisect the segment $\overline{MN}$ .

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