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.

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## Comments

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TopNewestNotice 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.

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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.

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@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.

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PS: The problem can be generalized.

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

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Yes. :)

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