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.

5 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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  4. Draw BD and BE; BD=BE
    ang(BDE)=ang(DEB)=a ; ang(EBG)=b => ang(FGB)=a+b => ang (FGC)=180-a-b
    ang(FAC)=angBAC)=ang(BDE)+ang(EBC)=a+b => ACGF is concyclic
    ang(GFC) and ang(GAC) on the same arc. => ang(GFC)=ang(GAC)

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  5. Let BT be the tangent in B of the circumcircle of ABC.
    B is the center of arc DE, therefore BT//DE
    Then, applying Reim’s reciprocal theorem, A,C,G,F are concyclic.

    (For this theorem, see my solution of problem n°77
    For the direct Reim’s theorem, see my solution of problem n°72)

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