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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.This entry contributed by Ajit Athle.Click the figure below to see the problem 769 details.
Partial Solution:Triangles AOM, COM are congruent,OE // AM, and OF // MC.So Triangles AEM,CFM have equal area.M lies on the bisector of angle ABC.So altitudes from M on bases AE and FC are of equal length.So AE = FC
total solution was sent to the email byMixalis Tsourakakis from Greece
Problem 769: Solution by Mixalis Tsourakakis from Greece. Thanks Mixalis.
To Mixalis TsourakakisI appreciate if you can translate your solution of problem 769 to English so that we can understand itPeter Tran
Because i don't speak english very fluently i only send you the text of the solution that Antonio Gutierrez attached translated in english without any symbols. The text is in the same order. My solution is very detailed and this is why is so lengthy. I hope this was helpful. Page 1PROOF USEFUL ELEMENTS FOR THE MAIN PROOFAs known (proof is very simple) it applies that ΝΒΜ= Let’s assume that Α>C in that case ΝΒΜ= [If angle Α< angle C then ΝΒΜ= ]ΗΝC+HXC=180 thus ΑΗΝ=CΑΗD=C+ΝΗD=C+ΝΒΜ (HD//BM)= C+ =S is the midpoint AC .MS mediator of ΑC (since ΑΜ=ΜC) and ΒCMS=Q .Then QA=QC and QAC=C.Also BAL=A-C and due to symmetry LCB=A.In addition MQ,AB,CL intersect at G.Page 2I define yellow triangles ΗDC ,OFC,OEG and blue triangles HAD,QOF,EOA. All yellow triangles are similar and all blue triangles are similar. Indeed: ΑΗD=AHN+NHD=C+NHD=C+NBM (HD//BM)= C+ = DHC=CHN-HDN=A- = (CHN=A because UHNA inscribable) and so HD is the bisector of ΑΗC . HAD=NBC(NABX inscribable)=(BN//QM)= OQF Since OF//MC, then Therefore OF is the bisector of triangle QOC. But QOF=OMC= (ABCM is inscribable and so ΑΜC=A+C). Thus also FOC= Since ΕΟ//ΑΜ,OF//MC then ΕΟF=AMC=A+C and since A+C+B=180 BEOF will be inscribable therefore ΟFC=BEO=EAM=A+ (CAM=ABM= ) It applies …………………………. Therefore OE is the bisector of triangle GOA.Page 3However GOE= thus also ΕΟΑ= Also it is obvious that G= HCD (G+A= A+HCD=90 G= HCD) Finally ΑΕΟ= ΒFO= (OF//MC) and ΗDC=BIC=A+ Therefore triangle ΑΗD triangle QOF since ΑΗD=QOF= and HAD= OQF Triangle QOF triangle EOA since EOA= QOF= and AEO=QFO= Thus all blue triangles are similar Triangle HDC triangle OFC since FOC= FOC=CHD and ΟFC=ΗDC=A+ Triangle HDC triangle GEO since G=UCA and DHC=GOE= Thus all yellow triangles are similar.MAIN PROOFSince ΗD is the bisector of triangle AHC it will be: Howerver triangle AHC triangle OQF and therefore triangle HDC triangle OFC and therefore Thus triangle FDC is isosceles, therefore DF=FC (1) triangle ΕΟΑ triangle HAD therefore triangle EOG triangle HDC thusThereforePage 4Therefore triangle ΑΕD is isosceles which means that ΕΑ=ΕD (2) For triangle ΕΖΑ applies ΑΕΖ=Γ+ , ΕΑΖ=Α-Γ therefore ΕΖΑ= Γ+ and hence triangle ΑΕΖ is isosceles and therefore ΕΑ=ΖΑ. But obviously ΖΑ=FC (3) From (1),(2),(3) DF=FC =ΕΑ=ΕDhttp://netload.in/dateiTp8iGM7rtV/Translation.rar.htmMichael Tsourakakis from Greece