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.


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

  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)

  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.

  4. < C = < E = < B

    Hence AEFB is cyclic and the result follows

    Sumith Peiris
    Sri Lanka