Competition problem — Geometry — Maths Olympiad Prep

Library / /93 of 120

Geometry Difficulty 6.2 National Olympiad Prove it Croatia

Let $ABCD$ be a trapezium with acute angles along the base $\overline{AB}$ and perpendicular diagonals which intersect at $O$ . Ray $OA$ intersects the circle with diameter $\overline{BD}$ at $M$ , and ray $OB$ intersects the circle with diameter $\overline{AC}$ at $N$ . Prove that the points $M$ , $N$ , $C$ and $D$ lie on the same circle. (New Zealand 2012)

Solution

Since $ABCD$ is a trapezium, the triangles $ABO$ and $CDO$ are similar and we have $\frac{|OA|}{|OB|} = \frac{|OC|}{|OD|}$ .

Euclid's theorem gives us $|OM|^2 = |OB| \cdot |OD|$ , $|ON|^2 = |OA| \cdot |OC|$ , so
$\frac{|OM|^2}{|ON|^2} = \frac{|OB| \cdot |OD|}{|OA| \cdot |OC|} = \frac{|OD|^2}{|OC|^2}.$
Now from $|OM| : |ON| = |OD| : |OC|$ and $\angle MON = \angle COD = 90^\circ$ we conclude that triangles $MON$ and $DOC$ are similar and hence $\angle MNO = \angle DCO$ .
It follows that $\angle MND = \angle DCM$ and therefore the points $C$ , $D$ , $M$ and $N$ are concyclic.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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