[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.

ABC\triangle ABC has a right angle at AA, DBCD\in BC and ADBCAD\perp BC. XX is an arbitrary point of ADAD. E,FCX,BXE,F\in CX,BX respectively such that AB=BE,AC=CFAB=BE,AC=CF.

Suppose BECF=GBE\cap CF=G. Prove that FG=GEFG=GE.

Extra Credit: Let XGBC=QXG\cap BC=Q. Prove that E,F,D,QE,F,D,Q are concyclic.

Note by Xuming Liang
3 years, 9 months ago

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Solution 1:

What is special about E,FE,F in relation to DD? Well, we do know from right triangle ABCABC that BE2=AB2=BDDC,CF2=AC2=CDCBBE^2=AB^2=BD\cdot DC, CF^2=AC^2=CD\cdot CB, implying BDEBEC,CDFCFBBDE\sim BEC, CDF\sim CFB.

Construct the orthocenter of XBCXBC denoted by HH. Then BED=XCB=BHD    B,H,D,E\angle BED=\angle XCB=\angle BHD\implies B,H,D,E are concyclic. Since BHCX,HEB=HDB=90BH\cap CX, \angle HEB=\angle HDB=90^{\circ}, we can easily deduce that HE2=HXHDHE^2=HX\cdot HD(try extending CXCX to meet BHBH). Likewise we can show HF2=HXHDHF^2=HX\cdot HD, so HE=HFHE=HF. Since HEG=HFG=90\angle HEG=\angle HFG=90^{\circ}, therefore GE=GFGE=GF.

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

Suppose DEFBE,BF=E,F\odot DEF\cap BE,BF=E',F'. Then EEQ=EDQ=XEB    XEEQ\angle EE'Q'=\angle EDQ'=\angle XEB\implies XE||E'Q'. Analogously QFFXQ'F'||FX. Furthurmore, FG=GEFG=GE implies that FEFEFEF'E' is an isosceles trapezoid. Hence EFEFEF||E'F', establishing that XEF,QEFXEF, Q'E'F' are homothetic and thus XQXQ' passes through the center of homothety which is GG. We are done.

Xuming Liang - 3 years, 9 months ago

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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

Surya Prakash - 3 years, 9 months ago

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

Xuming Liang - 3 years, 9 months ago

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

Nihar Mahajan - 3 years, 9 months ago

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Thanks @Xuming Liang Nice solution.

Surya Prakash - 3 years, 9 months ago

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

Xuming Liang - 3 years, 8 months ago

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