Sunday, December 26, 2010

Problem 556 Quadrilateral, Diagonal, Angles, Auxiliary Lines.

Geometry Problem
Click the figure below to see the complete problem 556.

 Problem 556. Quadrilateral, Diagonal, Angles, Auxiliary Lines.
Level: High School, SAT Prep, College geometry


  1. Bisect Angle BCD with CE (E is on AD), connect BE; bisect Angle EBD with BF (F is on AC), connect EF.
    Triangles BCE & DCE are congruent, Angle CBE=78 Degrees. Since Angles CBE and BCD are bisected, Angle EBF=Angle CBF=39 Degrees. Angle BCE=Angle DCE=42 Degrees. So AC bisects Angle BCE, Angle ACB=Angle ACE=21 Degrees. Thus, Triangle BCE's incenter is F (EF bisects Angle BEC, Angle BEF=Angle CEF=30 Degrees.
    ABEF is concyclic (two 39 Degree angles subtended on EF), therefore x=30 Degrees.

  2. A nice problem indeed!
    Draw an arc centre at C with radius CB cutting AD at E.
    CE = CD => AE = CE
    CB = CE, Triangle BCE is equilateral
    Follows AE = BE, <BAE (= <ABE) = 69 deg
    Hence x + 39 deg = 69 deg,
    x = 30 deg

  3. Nice problem
    two nice solutions