[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
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 $I'G,I'H$ 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.

- 4 years ago

@Xuming Liang Next in the series pls? :D

- 1 year, 2 months ago

nice and easy one.

- 4 years ago

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

- 1 year, 2 months ago