Thursday, February 5, 2009

Elearn Geometry Problem 245: Parallelogram with Equilateral triangles. Congruence

Problem: Parallelogram with Equilateral triangles. Congruence

See complete Problem 245 at:
gogeometry.com/problem/p245_parallelogram_equilateral_triangle_congruence.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. BE=AD=EC ; AB=DF=CF ; angle ABE = angle ADF = angle ECF , so the triangles ABE , FDA , FCE are congruents, so AE=AF=EF.

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  2. So what if i want to find the value of BEC or BCE angle?

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  3. DF=DC=AB
    AF=AE
    AD=BC=BE
    Triangle DAF congruent to triangle BEA
    So <BEA=<DAF=x
    Let AE and BC meets at G
    <BGA=60+x (ext < of triangle)
    <EAD=<BGA=60+x (alt <s, BC//AD)
    <EAF=<EAD-<DAF=60
    ---------------------------------------
    DF=FC
    DA=BC=CE
    FA=FE
    Triangle DAF congruent to triangle CEF
    So, <DFA=<CFE
    <CFD=<DFA+<CFA=60
    <CFA=60-<DFA=60-<CFE
    So, <CFE+<CFA=60=EFA
    So, AEF is an equilateral triangle since all 3 <s=60

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  4. Triangles ABE, ECF, ADF are congruent SAS
    So AE = EF = FA
    Hence Triangle AEF is equilateral

    Sumith Peiris
    Moratuwa
    Sri Lanka

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