Monday, May 19, 2008

Elearn Geometry Problem 33: Triangle and Quadrilateral

Triangle and Quadrilateral angles

See complete Problem 33 at:

Triangle and Quadrilateral angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. I have this question:

    If K is the circumcircle of BCD, why isnt direct that A is the center? I mean: angles BAC and BDC have same arc and one is half of the other...

    1. Consider the circumcircle of triangle BAD, then, arc BD is the locus of all P points such that <CPB=2<CDB

      Greetings :)

  2. Let B' is reflection of B in AD.
    Then, Angle AB'D=Angle DCE, so B,C,D,B' is cyclic.
    Also, Angle BDC=Angle BB'C=theta=Angle B'CA (because Angle BCD=2*theta) It means that AB'=AC. and BA=B'A by B' is reflection of B.
    So, A is center of circle passing B,C,D,B'. Therefore AD=AB, x=45 degree.

  3. Extend BA to F such that BA = AF. Then < DFA = x and hence BCDF is con cyclic. So < BFC = theta, hence < ACF = theta. So BA= CA= FA and hence < BCF = 90

    Now < FCD = < FBD = < DCE = x, hence 2x= 90 and x=45

    Sumith Peiris
    Sri Lanka

  4. This is true irrespective of the value of theta.