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See complete Problem 77Angles in a circle, Cyclic Quadrilateral, Parallel lines. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Join EF. AB is parallel to CD so angle BAE = angle CDE = angle CFE (angles in the same sector)Hence angle BAE = angle CFE. So AEFB is cyclic. Hence angle AEB = angle AFB (angles in the same sector)Ajit: email@example.com
construction : join EF angle DCF = angle DEF since they are in the same segment. similarly, we have angle CDE = angle CFD. ...(i)but we are given that AB is parallel to CD, so angle ABC = angle BCDangle BAD = angle ADC as they are alternate interior angles ...(ii)but angle AEF + angle DEF = 180 degreesand angle BFE + angle CFE = 180 degrees as they constitute linear pairs. ...(iii)using the results of (i), (ii) and (iii) we get that angle ABF + angle AEF = 180 degrees and angle BAE + angle BFE = 180 degreesso that ABFE is a cyclic quadrilateral the sum of the two pairs of opposite angles being 180 degrees.which implies that angle AEB = angle AFB.Q. E. D.
< C = < E = < B Hence AEFB is cyclic and the result followsSumith PeirisMoratuwaSri Lanka