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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 4.4 AIME Prove it Romania

In the acute-angled triangle $ABC$ , with $AB \neq AC$ , $D$ is the foot of the angle bisector of angle $A$ , and $E$ , $F$ are the feet of the altitudes from $B$ and $C$ , respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$ . Prove that $ME = MF$ .

Solution

Solution:

Triangles $AEF$ and $ABC$ are similar, therefore $AF \cdot AB = AE \cdot AC$ . It follows the point $A$ is on the radical axis of the two circumcircles, hence $M \in AD$ . We have that $m(\angle EMF) = 360^\circ - (180^\circ - m(\angle FBD)) - (180^\circ - m(\angle ECD)) = m(\angle B) + m(\angle C) = 180^\circ - m(\angle A)$ ; it follows that the quadrilateral $AEMF$ is cyclic. This means that $\angle MEF \equiv \angle FAM$ and $\angle MFE \equiv \angle EAM$ , i.e. triangle $MEF$ is isosceles, with $ME = MF$ .

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