Monday, November 7, 2016

Geometry Problem 1284 Two Equilateral Triangle, Midpoint, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1284.

Geometry Problem 1284: Two Equilateral Triangle, Midpoint, Congruence.

4 comments:

  1. https://goo.gl/photos/49eNJ5Yuiq4nrZ5P8
    Draw M on BF such that FM=FB ( see sketch)
    We have triangle ACM is equilateral
    Triangle ACE congruent to MCD ( case SAS)
    And triangle MCD is the image of ACE in the rotational transformation center at C, rot. Angle= 60 degrees.
    So AE=MD and angle(AE, MD)= 60
    And FH=1/2.AE and FH//AE
    FG=1/2.MD and FG//MD
    So FG=GH and angle(FG, FH)= angle (AE,MD)= 60 degrees
    And FGH is equilateral

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  2. Problem 1284
    Let the equilateral triangle AEP (PA,PE intersects the BD, not the extensions ). Then triangle ABP=triangle ACE (AB=AC,AP=AE, <BAP=60-<PAC=<CAE).So BP=CE=CD,but
    <PBC+<BCD=<ABP-60+<BCD=<ACE-60+<BCD=240-60=180, then BP//CD so the BPDC is
    Parallelogram .Then the points P,G and C are collinear with PG=GC.So GF=AP/2=AE/2=FH
    and <HFG=<EAP=60 (GF//AP,FH//AE).Therefore triangle FGH is equilateral.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  3. Complete the rhombus ABCX.

    Triangles ACE ≡ XCD, SAS.

    So AE = XD and hence FH = FG from the mid point theorem in triangles ACE
    and XBD.

    Similarly we can show that GH = FH by completing rhombus CDEY.

    Hence FG = FH = GH

    Sumith Peiris
    Moratuwa

    ReplyDelete
  4. Dear Antonio

    My solution does not appear yet, hence resenting.

    Complete the rhombus ABCX.

    Triangles ACE ≡ XCD, SAS.

    So AE = XD and hence FH = FG from the mid point theorem in triangles ACE
    and XBD.

    Similarly we can show that GH = FH by completing rhombus CDEY.

    Hence FG = FH = GH

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete