Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Belarus

Given a cyclic quadrilateral $ABCD$ with $AB = AD$ . Points $M$ and $N$ are marked on the sides $CD$ and $BC$ , respectively, so that $DM + BN = MN$ .

Prove that the circumcenter of the triangle $AMN$ belongs to the segment $AC$ .

Solution

On the prolongation of the segment $CD$ over $D$ we mark the point $K$ such that $DK = BN$ . Then the triangles $KDA$ and $NBA$ are equal since $\angle KDA = \angle NBA$ , $KD = BN$ , $DA = AB$ . Hence $KA = NA$ , $KM = DK + DM = BN + DM = NM$ , so that the triangles $KMA$ and $NMA$ are equal. It easily follows that $\angle MAN = 0.5\angle DAB$ . Let $O$ , $C$ be the intersection points of $AC$ and the circumcircle of the triangle $MCN$ .

Since $\angle MCO = \angle DCA = \angle BCA = \angle NCO$ , we have $MO = NO$ . Besides,
$\angle MON = 180^\{\circ\} - \angle MCN = 180^\{\circ\} - \angle DCB = \angle DAB = 2\angle MAN.$
This equality along with $MO = NO$ means that $O$ is the circumcenter of the triangle $MAN$ .

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