[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 \(\angle FHB=\angle EGC\)

#Geometry

Note by Xuming Liang
5 years, 8 months ago

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Comments

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′HI'G,I'H more so that they intersect AB,ACAB,AC, how can symmetry suggest what to do next?

Solution 1: Construct (FBH)∩CF=J≠F,(ECG)∩BE=K≠E(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 I′G,I′HI'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 XY∣∣KJXY||KJ. Note that ∠IGX=∠IBI′=∠B=∠CJH\angle IGX=\angle IBI'=\angle B=\angle CJH, thus IGX∼HJYIGX\sim HJY. Likewise we can show IHY∼GKXIHY\sim GKX. Hence by ratios, JY⋅IX=XG⋅HY=XK⋅IY  ⟹  JKIY=XKIX  ⟹  XY∣∣KJJY\cdot IX=XG\cdot HY=XK\cdot IY\implies \frac {JK}{IY}=\frac {XK}{IX}\implies XY||KJ. We are done.

Xuming Liang - 5 years, 8 months ago

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@Xuming Liang Next in the series pls? :D

Vishwash Kumar ΓΞΩ - 2 years, 10 months ago

nice and easy one.

Surya Prakash - 5 years, 8 months ago

Denote by I′G∩AB=K,I′H∩AC=L.I'G \cap AB= K, I'H \cap AC =L. Then, ∠AIF=∠AKI=90∘−∠B2\angle AIF = \angle AKI = 90^{\circ} - \angle \frac{B}{2} ∴\therefore By PoP, AF×AK=AI2AF \times AK = AI^{2} Similarly, AE×AL=AI2⇒EFKLAE \times AL = AI^{2} \Rightarrow EFKL is cyclic ⇒∠EKF=∠FLE⟹∠FHB=∠EGC.\Rightarrow \angle EKF = \angle FLE \Longrightarrow \angle FHB =\angle EGC. ■\blacksquare

Vishwash Kumar ΓΞΩ - 2 years, 10 months ago
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