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Proposed ProblemClick the figure below to see the complete problem 385.See also:Complete Problem 385Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
Draw FG the altitude of Tr. BFC. Since BF bisects < B, < FBD =90 - A/2 - B/2 = C/2 = < FED since BEFD is cyclic and is also = < ECA. Hence DE is parallel to ACSo < BED = 90 - C/2 = < EBG considering Tr. BEC. Since BEDG is cyclic it follows that the chords subtending these = angles are also =. Hence DE = BG = s-bSumith PeirisMoratuwaSri Lanka