Saturday, January 31, 2009

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

Problem: Square, Midpoints, Lines, Congruence

See complete Problem 238 at:
gogeometry.com/problem/p238_square_midpoints_congruence.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 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.

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  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: ajitathle@gmail.com

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  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.

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  4. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=32650&p=151226

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