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Geometry Difficulty 8.7 Shortlist Prove it Balkan Mathematical Olympiad

Let $ABC$ be a triangle with circumcircle $c$ and circumcenter $O$ , and let $D$ be a point on the side $BC$ different from the vertices and the midpoint of $BC$ . Let $K$ be the point where the circumcircle $c_1$ of the triangle $BOD$ intersects $c$ for the second time and let $Z$ be the point where $c_1$ meets the line $AB$ . Let $M$ be the point where the circumcircle $c_2$ of the triangle $COD$ intersects $c$ for the second time and let $E$ be the point where $c_2$ meets the line $AC$ . Finally let $N$ be the point where the circumcircle $c_3$ of the triangle $AEZ$ meets $c$ again. Prove that the triangles $ABC$ and $NKM$ are congruent.

Solution

Since the quadrilateral $ODCE$ is cyclic, we have $\angle O_1 = \angle C$ . (The angles are as shown in the figure.)

Since the quadrilateral $ODBZ$ is cyclic, we have $\angle O_2 = \angle B$ . Adding these we obtain $\angle EOUZ = \angle B + \angle C = 180^\circ - \angle A$ . Therefore the points $A$ , $Z$ , $O$ , $E$ are concyclic and $c_3$ passes through $O$ .
We also have $\angle Z_2 = \angle D_2 = \angle E_1$ . Since these three angles are subtended by the chords $AO$ , $BO$ , $CO$ of equal length in the circles $c_3$ , $c_1$ , $c_2$ , respectively, the radii of these circles are equal. Therefore the angles $\angle Z_1$ and $\angle D_1$ are equal as they are subtended by chords $OK$ and $OM$ of the same length. It follows that the points $K$ , $D$ , $M$ are collinear. Similarly, $M$ , $E$ , $N$ are collinear and $Z$ , $Z$ , $K$ are collinear. Then $\angle BDK = \angle CDM$ and $\angle CEM = \angle AEN$ , and $BK = CM = AN$ . Therefore the triangle $NKM$ is obtained by rotating the triangle $ABC$ by a rotation with center $O$ . Hence $ABC$ and $NKM$ are congruent.

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