TITLE: Macedonia National Olympiad 2010
QUESTION [5 upvotes]: Problem
The point O is the center of the circumscribed circle of the acute-angled triangle ABC. The line AO cuts the side BC in point N, and the line BO cuts the side AC at point M. Prove that if CM = CN, then AC = BC.
Attempt
Let AB = $c$, AC = $b$, BC = $a$
$\frac{NB}{NC} = \frac{S_{ANB}}{S_{ANC}}=\frac{c\space sin\angle{BAN}}{b\space sin\angle{CAN}} = \frac{c\space sin\space(\frac{\pi}{2}-\angle{BPN})}{b\space sin(\frac{\pi}{2}-\angle{CPN})} = \frac{c\space cos\angle BPA}{b\space cos\angle CPA} = \frac{c\space cos\angle C}{b\space cos\angle B}$
$\frac{a}{NC} = \frac{BC}{NC} = \frac{BN + NC}{NC} = \frac{c\space cos\angle C + b\space cos\angle B}{b\space cos\angle B}$
Similarly, 
$\frac{b}{MC} = \frac{c\space cos\angle C + a\space cos\angle A}{a\space cos\angle A}$
Because $CM = CN$, so
$\frac{c\space cos\angle C + a\space cos\angle A}{ab\space cos\angle A} = \frac{c\space cos\angle C + b\space cos\angle B}{ab\space cos\angle B}$ or $\frac{c\space cos\angle C + a\space cos\angle A}{cos\angle A} = \frac{c\space cos\angle C + b\space cos\angle B}{cos\angle B}$
However, I am not able to move forward from here, I admit that my ability to manipulate trig equations is relatively weak.

REPLY [2 votes]: First draw a line that pass through $C$ and $O$, and let that line intersects the circumcircle at $D$ and the side $\overline{AB}$ at $E$.
Now we'll prove that by contradiction. Let's $CM = CN$ and WLOG let $AC > BC$. It's obvious that $\angle ABC > \angle BAC$. This is equivalent to $\angle ADC > \angle BDC$ as angles above same arc.
From the Thales' Theorem we have that the inscribed angles $\angle DAC = \angle DBC = 90 ^{\circ}$, and obviously we have:
$$\angle ACE < \angle BCE \iff \angle AOE < \angle BOE$$
as central angles.
Now becasue the $\triangle ABO$ is obviously isoscelec we have that $AE < BE$
Now from Ceva Theorem we have:
$$\frac{AM}{MC} \times \frac{CN}{NB} \times \frac{BE}{EA} = 1$$
Because we assumed $$CN=CM$$ we have:
$$\frac{AM}{NB} = \frac{EA}{BE} < 1 \implies AM < NB$$
And at last we have:
$$AC = AM + CM = AM + CN < BN + CN = BC$$
But this is contradictory with out assumption. Hence the proof.