Wednesday, June 20, 2012

Problem 768: Parallelogram with Equilateral Triangles on Sides

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 768.

Online Geometry Problem 768: Parallelogram with Equilateral Triangles on Sides.

Posted by Antonio Gutierrez at 6:50 AM
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4 comments:

  1. Jacob HaJune 20, 2012 at 8:04 AM

    First, we have
    (1) DH = AD = BC = BF
    (2) DG = CD = AB = BE
    (3) Angle GDH = 240 - Angle ADC
    = 240 - Angle ABC = Angle EBF

    Hence, triangle DGH congruent to triangle BEF.
    Thus, GH = EF.

    Similarly, we can prove that FG = EH.
    Hence, EFGH is a parallelogram.

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  2. Peter TranJune 20, 2012 at 8:06 AM

    EB=BA=CD=DG
    And BF=BC=AD=DH
    And ∠(EBF)=360-120-∠(ABC)= ∠(GDH)
    ∆ EBF congruence to ∆ GDH ….( Case SAS) => EF=HG
    Similarly we have ∆EAH congruence to ∆ GCF => EH=FG
    So EFGH is a parallelogram

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  3. PravinJune 21, 2012 at 10:18 AM

    Let AC ∩ BD = O
    ∆OBF ≡ ∆ODH
    ∴ ∠BOF = ∠DOH and OF = OH
    F,O,H are collinear and FH is bisected at O.
    ///ly EG is bisected at O.
    Hence EFGH is a //ogram

    ReplyDelete
  4. Sumith PeirisSeptember 20, 2015 at 6:12 AM

    The opposite Tr. s thus formed by the equilateral triangles are congruent SAS since the opposite angles of the original parellelogram are equal.

    So the opposite sides of these Tr. s are equal. Hence the quadrilateral has opposite sides equal and is thus a parallelogram

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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