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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.8 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle, $D$ the midpoint of side $BC$ and $E$ the intersection point of the bisector of angle $\angle BAC$ with side $BC$ . The perpendicular bisector of $AE$ intersects the bisectors of angles $\angle CBA$ and $\angle CDA$ at $M$ and $N$ , respectively. The bisectors of angles $\angle CBA$ and $\angle CDA$ intersect at $P$ . Prove that points $A, M, N, P$ are concyclic.

Solution

The bisector of angle $\angle EBA$ and the perpendicular bisector of side $EA$ , both bisect the $\text{arc EA}$ of the circumcircle of triangle $ABE$ opposite to the vertex $B$ . Therefore quadrilateral $ABEM$ is cyclic and hence
$\angle EAM = \angle EBM$
The segment defined by the two intersection points of the perpendicular bisector of $AE$ with the circumcircle of triangle $AED$ is a diameter and intersects the bisector of the angle $\angle ADE$ on this circumcircle. But the bisectors of angles $\angle CDA$ and $\angle ADE$ are perpendicular. Therefore quadrilateral $AEDN$ is cyclic and hence
$\angle CDN = \angle EAN.$

We deduce that
$\angle MAN = \angle EAN - \angle EAM = \angle CDN - \angle EBM = \angle MPN,$
from triangle $BDP$ . This proves that points $A, M, N, P$ are concyclic.

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