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.

Posted by Antonio Gutierrez at 9:23 PM
Labels: , , ,

11 comments:

  1. macsim gabriel florinFebruary 2, 2009 at 3:08 AM

    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.

    ReplyDelete
  2. AjitApril 15, 2009 at 8:28 PM

    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

    ReplyDelete
  3. Ramprakash.KFebruary 5, 2011 at 9:38 AM

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

    ReplyDelete
  4. Μιχάλης ΝάννοςOctober 10, 2013 at 6:59 AM

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

    ReplyDelete
  5. Sumith PeirisJune 1, 2020 at 8:32 AM

    ABE,BCF,ADF are congruent right triangles.

    So < BAE = < FBC = < FAD = p say.

    AE & BF are hence perpendicualar and AGFD is concyclic and hence < GAF = 90-2p & so < GFA = 2p = < ADG

    Therefore < GAD = < AGD = 90-p and so b = GD = AD = a

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. GregSeptember 29, 2020 at 12:53 PM

    - b is the length of chord GD intercepted by angle <DAG in circle ADFG of diameter AF
    - a is the length of chord AB intercepted by angle <AEB in circle ABE of diameter AE
    -but AE = AF and <AEB = <DAG hence b = a

    ReplyDelete
  7. rv.littlemanSeptember 30, 2020 at 1:04 AM

    See the drawing

    - Triangle ABE is congruent to triangle BCF by rotation of pi/2 => < AGF=90°
    - Define H as the middle of AB and K as the middle of AG
    - By construction HD//BF
    - AH=HB=>AK=KG
    - < AGF=90° => < AKD=90°
    - triangle ADG has its height in the middle of AG
    - => ADG is isosceles in D => DA=DG
    - Therefore a=b

    ReplyDelete
  8. c.t.e.o.September 30, 2020 at 11:24 AM

    Draw DLM, L on AG, M midpoint of AB. Tr AML~ Tr GLD with coefficient 0.5

    ReplyDelete
  9. rv.littlemanOctober 7, 2020 at 7:14 AM

    See the drawing


    - ΔABE is congruent to ΔBCF by rotation of pi/2 => ∠ AGF=90°
    - Define H as the middle of AB and K as the middle of AG
    - By construction HD//BF
    - AH=HB=>AK=KG
    - ∠ AGF=90° => ∠ AKD=90°
    - ΔDKG is congruent to ΔDKA (SAS)
    - =>DG=DA
    - Therefore a=b

    ReplyDelete
  10. MarcoMarch 12, 2023 at 3:07 AM

    Triangle BGE ~ Triangle BCF
    BG:GE=2:1
    GE=c, BG=2c
    ----------------------------------------
    By Pyth. Theorem (Triangle BEG)
    BE^2=BG^2+GE^2=5c^2
    BE^2=(a/2)^2
    =>a^2=20c^2
    ----------------------------------------
    By Pyth. Theorem (Triangle BAE)
    BE^2+AB^2=AE^2
    AE^2=1.25a^2=25c^2
    AE=5c
    AG=AE-GE=4a
    -----------------------------------------
    cos0

    ReplyDelete
  11. Sumith PeirisMarch 12, 2023 at 9:02 AM

    Solution 2

    Triangles ABE, AFD & BCF are congruent SAS
    So < BFC = < AEB and hence EGCF is concyclic
    BF is thus perpendicular to AE and so ADFG is also concyclic
    Therefore < BFC = < GAD = < AFD = < AGD
    So AD = GD and a = b

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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