[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 ABC\triangle ABC, let BE,CFBE,CF be angle bisectors such that EAC,FABE\in AC, F\in AB. Reflect the incenter IIof ABCABC over BCBC to obtain II'. G,HBCG,H\in BC such that IGBE,IHCFI'G\perp BE, I'H\perp CF. Prove that FHB=EGC\angle FHB=\angle EGC

Note by Xuming Liang
4 years, 1 month 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 IG,IHI'G,I'H more so that they intersect AB,ACAB,AC, how can symmetry suggest what to do next?

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

Extend IG,IHI'G,I'H to meet BI,CIBI,CI at X,YX,Y. Since B,C,Y,XB,C,Y,X can be proven to be concyclic, it suffices to show that XYKJXY||KJ. Note that IGX=IBI=B=CJH\angle IGX=\angle IBI'=\angle B=\angle CJH, thus IGXHJYIGX\sim HJY. Likewise we can show IHYGKXIHY\sim GKX. Hence by ratios, JYIX=XGHY=XKIY    JKIY=XKIX    XYKJJY\cdot IX=XG\cdot HY=XK\cdot IY\implies \frac {JK}{IY}=\frac {XK}{IX}\implies XY||KJ. We are done.

Xuming Liang - 4 years ago

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@Xuming Liang Next in the series pls? :D

The almighty knows it all. - 1 year, 2 months ago

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nice and easy one.

Surya Prakash - 4 years ago

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Denote by IGAB=K,IHAC=L.I'G \cap AB= K, I'H \cap AC =L. Then, AIF=AKI=90B2\angle AIF = \angle AKI = 90^{\circ} - \angle \frac{B}{2} \therefore By PoP, AF×AK=AI2AF \times AK = AI^{2} Similarly, AE×AL=AI2EFKLAE \times AL = AI^{2} \Rightarrow EFKL is cyclic EKF=FLEFHB=EGC.\Rightarrow \angle EKF = \angle FLE \Longrightarrow \angle FHB =\angle EGC. \blacksquare

The almighty knows it all. - 1 year, 2 months ago

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