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

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Thanks @Xuming Liang Nice solution.

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

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