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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.2 AIME, harder Prove it Saudi Arabia

Let $ABC$ be an acute triangle with $\angle A < \angle B \leq \angle C$ , and $O$ its circumcenter. The perpendicular bisector of side $AB$ intersects side $AC$ at $D$ . The perpendicular bisector of side $AC$ intersects side $AB$ at $E$ . Express the angles of triangle $DEO$ in terms of the angles of triangle $ABC$ .

Solution

Let $M$ be the midpoint of $AC$ . Because $\measuredangle COM = \frac{1}{2} \measuredangle COA = \measuredangle CBE$ , the quadrilateral $BCOE$ is cyclic.

Let $N$ be the midpoint of $AB$ . Because $\measuredangle NOB = \frac{1}{2} \measuredangle AOB = \measuredangle DCB$ , the quadrilateral $BCDO$ is cyclic. Therefore, the pentagon $BCDOE$ is cyclic and we have:
$\measuredangle ODE = \measuredangle OBE = 90^{\circ} - \measuredangle ACB, \quad \measuredangle DEO = \measuredangle DCO = 90^{\circ} - \measuredangle CBA,$
and
$\measuredangle EOD = 180^{\circ} - \measuredangle ODE - \measuredangle DEO = \measuredangle CBA + \measuredangle ACB.$

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