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# 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
2 years 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.

- 2 years 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.

- 2 years 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.

- 2 years ago

PS: The problem can be generalized.

- 2 years ago

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

- 2 years ago

Yes. :)

- 2 years ago