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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 627.
Denote each of the equal angles ABD, ACD by xAB = BE, CD=CF imply that each of the angles AEB, CFD is (180° - x)/2Taking supplements,angle AED = angle AFDSo A,E,F,D are concyclic
The internal bisectors of the angles ABD and ACDmeet at the midpoint O of the (minor)arc AD.Also O lies on the perpendicular bisector of AE as well as FD.So OE = OA = OD = OF = k say∴A, E, F, D lie on the circle (O;k)
Since < ABE = < ACD, Tr.s ABE & FCD are similar, hence < AED = < AFD and the result follows.Sumith PeirisMoratuwa Sri Lanka