Tuesday, August 6, 2024

Geometry Problem 1576: Congruency of Segments in Triangle ABC with Angles 30 and 20 Degrees and an Interior Cevian

Challenging Geometry Puzzle: Problem 1576. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1576

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Posted by Antonio Gutierrez at 3:02 PM
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4 comments:

  1. Sumith PeirisAugust 7, 2024 at 12:12 AM

    Trigonometry Solution

    Let BD = CD = p and let AD = d

    From Triangle ABD,
    p/sin30 = d/sin70 ................(1)

    From Triangle BCD,
    a = 2p cos20 = 2p sin70 ............(2)

    (1) X (2) gives,
    ap/sin30 = 2pd
    So a = 2dsin30 = d

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Sumith PeirisAugust 7, 2024 at 3:54 AM

    Pure Geometry Solution

    Extend BD to E such that BD = DE, let BD = DC = DE = p
    Let DF be perpendicular to AB, F on AB extended
    Let AD = d

    Note that ADF is a 30-60-90 Triangle and so DF = d/2

    Now < BCE = 90 and Triangles BFD and BCE are similar

    Hence DF / BD = BC / BE which gives
    (d/2)/p = a/ (2p)

    Therefore d = a, that is AD = BC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Stan FULGEROctober 13, 2024 at 9:45 AM

    My synthetic solution on my blog, https://stanfulger.blogspot.com/2024/10/prob-1576-gogeometry.html

    ReplyDelete

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