I used to think algebraic geometry and analytic geometry are the same thing. How naive was I?

Given \(\triangle ABC\), let \(BE,CF\) be angle bisectors such that \(E\in AC, F\in AB\). Reflect the incenter \(I\)of \(ABC\) over \(BC\) to obtain \(I'\). \(G,H\in BC\) such that \(I'G\perp BE, I'H\perp CF\). Prove that \(\angle FHB=\angle EGC\)

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## Comments

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TopNewestI came up with this solution with the intention to use cyclic, therefore it is different than my original solution(3 years old). Instead of posting that as solution 2, I will give out hints to allow someone else to devise the proof: Extend \(I'G,I'H\) more so that they intersect \(AB,AC\), how can symmetry suggest what to do next?

Solution 1: Construct \((FBH)\cap CF=J\ne F, (ECG)\cap BE=K\ne E\). Then \(\angle FHG=\angle EGC\iff \angle FJB=\angle EKC\iff B,C,K,J\) are concyclic.

Extend \(I'G,I'H\) to meet \(BI,CI\) at \(X,Y\). Since \(B,C,Y,X\) can be proven to be concyclic, it suffices to show that \(XY||KJ\). Note that \(\angle IGX=\angle IBI'=\angle B=\angle CJH\), thus \(IGX\sim HJY\). Likewise we can show \(IHY\sim GKX\). Hence by ratios, \(JY\cdot IX=XG\cdot HY=XK\cdot IY\implies \frac {JK}{IY}=\frac {XK}{IX}\implies XY||KJ\). We are done.

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nice and easy one.

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Comment deleted Sep 20, 2015

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no, u r wrong.

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I meant that if that is trye the problem will be true. I am not sure but am working on the problem.

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Comment deleted Sep 20, 2015

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