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NMO Selection Tests for JBMO

Geometry Difficulty 5.9 AIME, harder Prove it Romania

Let $\triangle ABC$ be an acute triangle, with $AB \neq AC$ . Let $D$ be the midpoint of the line segment $BC$ , and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$ , respectively. If $M$ is the midpoint of the line segment $EF$ , and $O$ is the circumcenter of triangle $ABC$ , prove that the lines $DM$ and $AO$ are parallel.

Caucasus Mathematical Olympiad

Solution

Let $S$ be the intersection point of the lines $AO$ and $BC$ . Let $T$ be the point in which the line $AD$ meets again the circumcircle of triangle $ABC$ . The quadrilaterals $ABTC$ and $AEDF$ are cyclic, hence $\angle TBD = \angle TAC = \angle DEF$ and $\angle TCD = \angle TAB = \angle DFE$ . It follows that the triangle $DEF$ and $TBC$ are similar. Then we have $\frac{TB}{DE} = \frac{BC}{EF} = \frac{BC/2}{EF/2} = \frac{BD}{EM}$ . This shows that the triangles $TBD$ and $DEM$ are also similar, hence $\angle EDM = \angle BTD = \angle ACB$ , which leads to $\angle BDM = \angle BDE + \angle EDM = 90^\circ - \angle ABC + \angle ACB$ . From $\angle OAC = 90^\circ - \angle ABC$ , it follows that $\angle ASB = \angle SAC + \angle ACB = 90^\circ - \angle ABC + \angle ACB = \angle MDB$ , which means that $AS$ is parallel to $MD$ .

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