Saturday, January 31, 2009

Elearn Geometry Problem 238: Square, Midpoints, Lines, Congruence

Problem: Square, Midpoints, Lines, Congruence

See complete Problem 238 at:

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. It is easy to prove that the measure of AGF is 90(traingles ABE and BCF are congruents). Let H be the intersection of BF and AD. FD= AB/2, FD is parallel to AB then, AD=DH, so GD is median of ortogonal triangle AGH. So GD = AH /2 = AD = AB.
    There is another solution with calculus.

  2. Here's another by analytical geom.: Let A be(0,0),B:(0,a),C:(a,a)& D:(a,0. So E:(a/2,a) & F:(a,a/2)and BF:x/2a + y/a = 1, AE:y = 2x. This makes G:(2a/5,4a/5) and GD^2=b^2=(2a/5-a)^2+(4a/5)^2 = a^2. Hence a=b.
    I want to know how this can be solved by calculus. Ajit:

  3. We observe that AGFD is a cyclic quadrilateral.
    So, angle GFC = GAD.
    also, angle BFC = AFD and AFD = AGD.
    => GAD = AGD. => GD = AD = a.