Triangle ABC inscribed in Circle O


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.


Let AB = $c$, AC = $b$, BC = $a$

$\frac{NB}{NC} = \frac{S_{ANB}}{S_{ANC}}=\frac{c\space sin\angle{BAN}}{b\space sin\angle{CAN}} = \frac{c\space sin\space(\frac{\pi}{2}-\angle{BPN})}{b\space sin(\frac{\pi}{2}-\angle{CPN})} = \frac{c\space cos\angle BPA}{b\space cos\angle CPA} = \frac{c\space cos\angle C}{b\space cos\angle B}$

$\frac{a}{NC} = \frac{BC}{NC} = \frac{BN + NC}{NC} = \frac{c\space cos\angle C + b\space cos\angle B}{b\space cos\angle B}$


$\frac{b}{MC} = \frac{c\space cos\angle C + a\space cos\angle A}{a\space cos\angle A}$

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.


1 Answer 1


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.


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