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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 622.
tr BEC isosceles => ang CEB= ang ECB=atr CAF isosceles => ang CFA= ang ACF=athus AEFB is cyclic.ACBD is a deltoid so ang D = ang C and therefore ang AEB+ ang ADB= ang AEB+ ang ACB=180 => AEBD is cyclic.finally AEFBD is cyclic.
Tr.s ACB and ADB are congruent hence < ADB = < ACB = < AFB since AC = AF so ADFB is cyclic. Similarly ADBE is cyclicSo AEFBD is cyclicSumith PeirisMoratuwaSri Lanka