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Macedonian Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it North Macedonia

The point $O$ is the centre of the circumcircle of triangle $\Delta ABC$ .
The line $AO$ intersects the side $BC$ in point $N$ , and the line $BO$ the side $AC$ in point $M$ . Prove that, if $CM = CN$ , then $AC = BC$ .

Solution

Then $\angle ABC > \angle BAC$ i.e. $\angle ADC = \angle ABC > \angle BAC = \angle BDC$ .
This implies

But
$\angle AOE = 2\angle ACE \text{ and } \angle BOE = 2\angle BCE.$
Hence
$\angle AOE < \angle BOE$ i.e. $\overline{AE} < \overline{BE}$
( $\triangle ABO$ is isosceles).
From Ceva's theorem we have:
$\frac{\overline{AM}}{MC} \cdot \frac{\overline{CN}}{NB} \cdot \frac{\overline{BE}}{EA} = 1,$
and it holds that $\overline{CM} = \overline{CN}$ (by condition).
We get that $\frac{\overline{AM}}{BN} = \frac{\overline{EA}}{\overline{BE}} < 1$ i.e. we have
$\overline{AC} = \overline{AM} + \overline{CM} = \overline{AM} + \overline{CN} < \overline{BN} + \overline{CN} = \overline{BC}$
which is in contradiction with (1).
Analogously we can show that $\overline{AC} < \overline{BC}$ is impossible. This implies $\overline{AC} = \overline{BC}$ .

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