Monday, May 19, 2008

Elearn Geometry Problem 77



See complete Problem 77
Angles in a circle, Cyclic Quadrilateral, Parallel lines. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. ∠FEA=ext∠DCB=ext∠ABF
    A,E,F,B concyc.

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  2. 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: ajitathle@gmail.com

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  3. 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 BCD
    angle BAD = angle ADC as they are alternate interior angles ...(ii)
    but angle AEF + angle DEF = 180 degrees
    and 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 degrees
    so 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.

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  4. < C = < E = < B

    Hence AEFB is cyclic and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. This picture can be considered the reciprocal of that of problem n° 72, where Reim’s direct therorem applies.
    Therefore the reciprocal of Reim’s theorem can be applied here, and since AB//CD, therefore A, B, E, F belong to a circle O’ intersecting the given circle on E and F.
    A, B, E, F being cocyclic, therefore <AEB=<AFB. q.e.d.
    For a full development of Reim’s theorem, especially the many variety of configurations where it applies, read (in french): http://jl.ayme.pagesperso-orange.fr/apropos.html

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  6. Let the intersection of AD & BC be P
    ABFC is a cyclic quad.
    => alpha=beta (< in same seg.)

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