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# [Solutions Posted]Geometry Proof Problem of the day #4

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 $$\angle FHB=\angle EGC$$

Note by Xuming Liang
1 year, 4 months ago

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I 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. · 1 year, 3 months ago

nice and easy one. · 1 year, 3 months ago

Comment deleted Sep 20, 2015

no, u r wrong. · 1 year, 4 months ago

I meant that if that is trye the problem will be true. I am not sure but am working on the problem. · 1 year, 4 months ago

nice!!! did u get any clues? · 1 year, 4 months ago

Comment deleted Sep 20, 2015