Saturday, July 20, 2024

Challenging Geometry Puzzle: Problem 1566. 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 1566

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4 comments:

  1. We show that triangles ADC and BEC are congruent.
    Indeed, the quadrilateral BEFD is concyclic because <EBD = external <DFC = 60º, then also <BEF = external <FDC = <ADC.
    So triangles ADC and BEC have two equal angles <EBC = EBD = 60º = <ACD, and <ADC = <FDC = <BEF = <BEC. Also these triangles have congruent side AC = BC, both opposite to the equal angles <ADC = <BEC (then congruence by ASA). Observe that AD = EC, both opposite to the 60º angles, and finally CD = BE.
    But AE + EB = AB = AC and CD = BE = EB implies AC = AE + CD, as desired.

    Joaquim Maia
    Rio de Janeiro
    Brazil

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  2. BEFD is concyclic since < AFD = 60 = < B
    So < BEC = < ADC
    Hence Triangles ADC & BEC are congruent ASA

    Therefore AE = BD = BC - CD = AC - CD from which
    AE + CD = AC

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. TrAEC,TrADB are congruent (AC=AC, <B=<B, <ACE=<BAD=<60°- <FAC) so
    AE=BD thus AE+CD=BD+CD=BC=AC

    Georgios Kousinioris
    Gastouni
    Greece

    ReplyDelete
  4. TrAEC,TrADB are congruent (AC=AB, <A=<B, <ACE=<BAD=<60°-<FAC) so
    AE=BD
    Therefore AE+CD=BD+CD=BC=AC

    Georgios Kousinioris
    Gastouni
    Greece

    ReplyDelete