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# [Solutions Posted]Geometry Proof Problem of the day #2 (diagram included)

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.

Note by Xuming Liang
1 year, 1 month ago

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Solution 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. · 1 year, 1 month ago

The first part of this problem turns out is from IMO 2012. · 1 year, 1 month ago

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 · 1 year, 1 month ago

Wish granted! Next time you could tag my name to get my attention faster. · 1 year, 1 month ago

Thanks @Xuming Liang Nice solution. · 1 year, 1 month ago