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 \(\angle FHB=\angle EGC\)

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

@Xuming Liang Next in the series pls? :D

Log in to reply

nice and easy one.

Log in to reply

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 \(\Rightarrow \angle EKF = \angle FLE \Longrightarrow \angle FHB =\angle EGC.\) \(\blacksquare\)

Log in to reply

Comment deleted Sep 20, 2015

Log in to reply

no, u r wrong.

Log in to reply

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

Log in to reply

Log in to reply

Comment deleted Sep 20, 2015

Log in to reply

Log in to reply