# A cool property regarding circumcenter

Let $O$ be the circumcenter of $ABC$. Reflect $O$ over $BC$ to obtain $O'$. Through $O'$ construct lines parallel to $AC,AB$ which respectively meet $AB,AC$ at $F,E$. Define $O'F\cap OB=Y, O'E\cap CO=X$. Prove $XY||EF$

I personally think this configuration is very rich and can be exploited to create difficult olympiad geo problems.

Note by Xuming Liang
4 years, 7 months ago

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Notice that $BOCO'$ is a parallelogram. Since $BP || O'C$ and $AB || O'E$, we see that $\angle FBX = \angle EO'C$. It is easy to see that $\angle A = \angle BFX = \angle YEC$. Therefore, $\triangle BFX \sim \triangle O'EC$. This implies that $\frac{BF}{FX} = \frac{O'E}{EC}$.

Similarly, we have that $\triangle YCE \sim \triangle BO'F \implies \frac{O'F}{FB} = \frac{CE}{EY}$.

Multiplying the two ratios completes the proof.

- 4 years, 7 months ago

Yes! The parallel property will hold as long as $BOCO'$ is a parallelogram. There are a couple of typos in your proof, otherwise you got it spot on.

- 4 years, 7 months ago

PS: The problem can be generalized.

- 4 years, 7 months ago

Can you post the proof or hint after a few days?

- 4 years, 7 months ago

Yes. :)

- 4 years, 7 months ago

@Alan Yan Hints: We want to prove some two ratios are equal, perhaps look for some similar triangles from all the parallels.

Generalize this property if you get it.

- 4 years, 7 months ago