Monday, May 19, 2008

Elearn Geometry Problem 67



See complete Problem 67
Triangle, Circumcircle, Angles, Concyclic. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. 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 BGF
    That is angle angle FAC = angle BGF
    Hence quadrilateral FACG is cyclic.
    Hence A, F, G and C are concyclic.

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  2. 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).

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  3. 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.

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