It's been a day or two? I did not keep track. Found this one in an online conversation I had with a geometer friend of mine; it might be or based on an existing olympiad problem though. I never keep track anyway.

\(\triangle ABC\) has a right angle at \(A\), \(D\in BC\) and \(AD\perp BC\). \(X\) is an arbitrary point of \(AD\). \(E,F\in CX,BX\) respectively such that \(AB=BE,AC=CF\).

Suppose \(BE\cap CF=G\). Prove that \(FG=GE\).

Extra Credit: Let \(XG\cap BC=Q\). Prove that \(E,F,D,Q\) are concyclic.

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## Comments

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TopNewestSolution 1:

What is special about \(E,F\) in relation to \(D\)? Well, we do know from right triangle \(ABC\) that \(BE^2=AB^2=BD\cdot DC, CF^2=AC^2=CD\cdot CB\), implying \(BDE\sim BEC, CDF\sim CFB\).

Construct the orthocenter of \(XBC\) denoted by \(H\). Then \(\angle BED=\angle XCB=\angle BHD\implies B,H,D,E\) are concyclic. Since \(BH\cap CX, \angle HEB=\angle HDB=90^{\circ}\), we can easily deduce that \(HE^2=HX\cdot HD\)(try extending \(CX\) to meet \(BH\)). Likewise we can show \(HF^2=HX\cdot HD\), so \(HE=HF\). Since \(\angle HEG=\angle HFG=90^{\circ}\), therefore \(GE=GF\).

Extra Credit: Let \(\odot DEF\cap BC=D,Q'\), we will show that \(Q'\in XG\) which implies \(Q\equiv Q'\).

Suppose \(\odot DEF\cap BE,BF=E',F'\). Then \(\angle EE'Q'=\angle EDQ'=\angle XEB\implies XE||E'Q'\). Analogously \(Q'F'||FX\). Furthurmore, \(FG=GE\) implies that \(FEF'E'\) is an isosceles trapezoid. Hence \(EF||E'F'\), establishing that \(XEF, Q'E'F'\) are homothetic and thus \(XQ'\) passes through the center of homothety which is \(G\). We are done.

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The first part of this problem turns out is from IMO 2012.

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Hey Xuming, please upload the solution. I am very eager to know about the proof. I tried alot but cant succeed. Atleast any HINT. :D

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Wish granted! Next time you could tag my name to get my attention faster.

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Thanks @Xuming Liang Nice solution.

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Nice one by revealing the hidden orthic configuration! :) :) :)

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