Tuesday, August 26, 2008

Elearn Geometry Problem 164

See complete Problem 164 at:

Parallelogram, Trapezoid, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Tr. ABD = Tr. BDC since AD=BC and the height is common. So S + Tr. ABF = S1 + S2 + Tr. DEF or S = S1 + S2 since Tr. ABF = Tr. DEF which have the same base and height.
    Ajit: ajitathle@gmail.com

  2. Akram Yousef BraiweshFebruary 1, 2010 at 12:42 AM

    s+tr EFD=1/2 ABCD
    s1+s2+tr EFD=1/2 ABCD
    So s+tr EFD=s1+s2+tr EFD wich means s=s1+s2

  3. If point E can be any point on BC, then let say that point E is on point B, meaning segment AE = segment AB, point F = point B, the area of S1 = 0 and the area of S = tr. ABD. This means that S = S2 because tr. ABD = tr. BDC. So, S = S1 + S2, also written as S = 0 + S2 since S1 = 0.

  4. The "solution" proposed by Anonymous, June 20, 2011 is not valid. He only proved that S = S1 + S2 in the case that E and B are the same point. What happens in the general situation?