Monday, August 7, 2023

Geometry Problem 1550: Solving for Segment BD: An Angle Puzzle in Right Triangle ABC

Challenging Geometry Puzzle: Problem 1550. 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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Geometry Problem 1550: Solving for Segment BD: An Angle Puzzle in Right Triangle ABC

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Posted by Antonio Gutierrez at 11:19 AM
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8 comments:

  1. Sumith PeirisAugust 8, 2023 at 12:31 PM

    Let AD & AE extended meet circle ABC (with diameter AC) at P & Q respectively.
    Draw DM perpendicular to BP, M on BP

    BP = CQ (both subtend 20 degrees on the circumference) = y, say

    < ACB = < DBP = < DPB = 35 and so,
    Triangle BDP is isosceles and hence BM = y/2

    Triangles BDM & CEQ are similar (90,35,55) and so
    BD/BM = CE/CQ

    Therefore BD/(y/2) = 3/y which yields
    BD = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Sumith PeirisAugust 8, 2023 at 11:40 PM

    Trigonometry Solution

    Let tan35 = t

    BD = c.tan20 = c/tan70 = c./(tan(2X35) =c.(1-t^2)/2t........(1)
    BE = c.tan35 = ct.......(2)
    BC = c.tan55 = c/tan35 = c/t .......(3)
    (3) - (2); CE = BC - BE = c.(1/t - t) from (2) & (3)
    So c.(1-t^2)/t = 3......(4)
    From (1) & (4) BD = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. CiPAugust 28, 2023 at 8:03 AM

    The trigonometric solution is beautiful because it is simple. The geometric solution is beautiful because it makes ingenious auxiliary constructions.

    ReplyDelete
  4. Sumith PeirisAugust 28, 2023 at 11:18 PM

    2nd Geometry Solution - much simpler

    Let FA be perpendicular to AC, F on CB extended
    Let BD = x

    Now FD = DA = DC so DE = a - x - 3 and FB = a - 2x
    FB = BE = a - 2x = BD + DE = x + (a - 3 - x) = a - 3

    Hence 2x = 3 and x = 3/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Rogerio de SouzaSeptember 10, 2023 at 6:10 PM

    Nice solution by creating the right triangle FAC.

    ReplyDelete
  6. Stan FULGERDecember 4, 2023 at 10:14 PM

    Notice that AD=CD and, if ADCF is a rhombus, F is the circumcenter of triangle AEC, <FCE=<ADB, FC||=AD, G, projection of F onto CB is midpoint of EC, while triangles CFG and DAB are congruent, so CG=BD, done.

    ReplyDelete

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