Sunday, August 11, 2024

Geometry Problem 1577: Prove AB Equals BD Plus CD in Triangle ABC with Given Angles 20, 30, 40 Degrees and Congruent Segments

Challenging Geometry Puzzle: Problem 1577. 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 1577

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Posted by Antonio Gutierrez at 2:20 PM
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1 comment:

  1. Sumith PeirisAugust 12, 2024 at 4:34 AM

    Extend AC upto E such that DC = CE = p say
    Let BD extended meet AC at F

    < CDE = < DEC = 40/2 =20 = < DBC
    So BDCE is concyclic

    Reference my Proof of Problem 1576,
    AC = DE = BD = b
    So < BED = 40 & < BEA = 60 = < FDC giving
    < BFC = < BCF = 80

    Hence Triangles BFC and DFE are both 80-20-80 Triangles with
    BF = BC and
    DE = FE = b
    So CF = b - p and so
    AF = p

    So in Triangles ABF & BCE
    AF = CE = p
    BF = BC and
    < BFA = < BCE = 100

    So Triangles ABF and CBE are congruent SAS and
    AB = AE
    Since < AEC = 60 it follows that Triangle ABE is equilateral

    Hence AB = AE = b + p = AC + CE and hence
    AB = BD + CD

    Sumith Peiris
    Moratuwa
    Sri Lanka

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