Wednesday, June 22, 2011

Problem 627: Cyclic Quadrilateral, Congruence, Concyclic Points

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 627.

 Problem 627: Cyclic Quadrilateral, Congruence, Concyclic Points.

3 comments:

  1. Denote each of the equal angles ABD, ACD by x

    AB = BE, CD=CF imply that

    each of the angles AEB, CFD is (180° - x)/2

    Taking supplements,

    angle AED = angle AFD

    So A,E,F,D are concyclic

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  2. The internal bisectors of the angles ABD and ACD
    meet at the midpoint O of the (minor)arc AD.
    Also O lies on the perpendicular bisector of AE as well as FD.
    So OE = OA = OD = OF = k say
    ∴A, E, F, D lie on the circle (O;k)

    ReplyDelete
  3. Since < ABE = < ACD, Tr.s ABE & FCD are similar, hence < AED = < AFD and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete