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.

Gain comprehensive insights! Click below to reveal the complete details.

Illustration of Geometry Problem 1576

Click for additional details.

Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.

Posted by Antonio Gutierrez at 3:02 PM
Labels: , , , ,

2 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

    ReplyDelete
  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

Subscribe to: Post Comments (Atom)