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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Notice that BOCO′ is a parallelogram. Since BP∣∣O′C and AB∣∣O′E, we see that ∠FBX=∠EO′C. It is easy to see that ∠A=∠BFX=∠YEC. Therefore, △BFX∼△O′EC. This implies that FXBF=ECO′E.
Similarly, we have that △YCE∼△BO′F⟹FBO′F=EYCE.
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.
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. :)
@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.