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See complete Problem 67Triangle, Circumcircle, Angles, Concyclic. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Join B and E. Join A and E.angle BAE = angle BED (angles subtended by equal arcs)angle EAC = angle CBE (angles subtended by the same arc)Hence angle BAE + angle EAC = angle BED + angle CBE But angle BED + angle CBE = angle BGF (exterior angle of triangle GBE = sum of interior opposite angles)Hence angle BAE + angle EAC = angle BGFThat is angle angle FAC = angle BGFHence quadrilateral FACG is cyclic.Hence A, F, G and C are concyclic.
Solution to Problem 67. In the circle circunscribed to ABC, ang(AFC) = (arc AC + arc BD)/2 and ang(AGC) = (arc AC + arc BE)/2. But arc BD = arc BE, so ang(AFC) = ang(AGC). These two angles subtend the same arc AC, thus AFGC is cyclic. In the circle circunscribed to AFGC, we have ang(CAG) = ang(CFG) because both subtend the same arc(CG).
Let PB be the tangent at B to the circumcircle of ΔABC(such that P, D lie on the same side of AB). Join BD, BE.∠PBD =∠BED (angle in the alternate segment)=∠BDE (since BD = BE) So PB ∥ DE.Now ∠BFG =∠PBF (alternate angles)=∠PBA =∠BCA (angle in the alternate segment)Hence AFGC is a cyclic quadrilateral etc.