Tuesday, February 21, 2023

Geometry Problem 1509: Congruence of Triangles in a Trapezoid and a Square, Measurement. Difficulty Level: High School

Geometry Problem 1509. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Geometry Problem 1509: Congruence of Triangles in a Trapezoid and a Square, Measurement. Difficulty Level: High School.

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Posted by Antonio Gutierrez at 6:44 PM
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13 comments:

  1. Pradyumna AgasheFebruary 21, 2023 at 8:35 PM

    Extend FB to point G such that BF=BG. Then we get CF=CG=CD, hence C is center of circle passing through points D,F and G. Since Angle DCF=90, Angle DGF=90/2=45.
    Hence Triangle ADG is right isosceles. We get AD=AG=AF+BF+BG=2+5+5=12.

    ReplyDelete
    Replies
    1. Sumith PeirisFebruary 22, 2023 at 9:33 AM

      Very elegant solution Pradyumna, much better than my own

      Delete
    2. Pradyumna AgasheFebruary 23, 2023 at 5:25 AM

      Thank you Sumith for your kind comment.

      Delete
  2. Sumith PeirisFebruary 22, 2023 at 5:05 AM

    Extend CF to G
    Let the side of the square be a, let GF = t and AG = q

    From similar triangles BCF, AGF and GCD
    5/a = 2/t = a/(q+x).....(1)

    Since AFCD is concyclic, t(a+t) = q(q+x)....(2)
    Substituting from (1) (2a/5)(a+2a/5) = (a^2/ 5 - x)(a^2/ 5)
    14/5 = a^2/ 5 - x and a^2 = 14 + 5x...(3)

    But x^2 +2^2 = 2.a^2
    Therefore x^2 + 4 = 28 + 10x from (3)
    x^2 - 10x - 24 = 0
    (x+2)(x-12) = 0

    So x = 12 since x = -2 is not permissible

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Peter TranFebruary 22, 2023 at 11:14 AM

    https://photos.app.goo.gl/tM61SHQ3HgmUEZrp9

    Let H is the project of C over AD
    Note that triangle FBC congruence to DHC ( case AA)
    So HD=BF=5 and CH=AB=7
    So AD=12

    ReplyDelete
  4. AnonymousFebruary 22, 2023 at 11:21 AM

    c.t.e.o.https://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9

    ReplyDelete
  5. c.t.e.o.February 22, 2023 at 11:22 AM

    https://photos.app.goo.gl/4E2MnjC9GN6Xd7fz9

    ReplyDelete
  6. Sumith PeirisMay 7, 2023 at 2:22 AM

    Another Solution.
    Complete Rectangle ADYX.
    Then Triangle BCF, XFE and YED are congruent ASA.
    So XE = DY = FB = 5 and hence, FX = EY = 7. Therefore AD = XE + EY = 5 + 7 = 12

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisMay 23, 2025 at 10:49 AM

    Let EG be perpendicular to AG, G on FA extended
    Let the square be of side a

    Triangles BCF and GFE are congruent ASA and so GE = 5
    Now AFDE is concyclic and so < GAE = < FDE = 45
    So Triangle GAE is Right Isoceles and AE = 5V2

    Now apply Ptolemy to AFDE
    ax = 5V2 X V2.a + 2a = 12a
    Therefore x = 12

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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