Tuesday, February 3, 2009

Elearn Geometry Problem 242: Triangle with Equilateral triangles, Parallelogram

Problem: Triangle with Equilateral triangles, Parallelogram

See complete Problem 242 at:
gogeometry.com/problem/p242_triangle_equilateral_parallelogram.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. C'A = BA
    angle C'AB' = Angle BAC = 60 - angle B'AB
    AB' = AC
    So, by SAS postulate,
    triangle C'AB' is congruent to triangle BAC
    So, C'B' = BC = BA'
    Similarly as
    triangle A'B'C is congruent to triangle BAC,
    A'B' = BA = BC'

    So,
    in quadrilateral A'BC'B' both pairs of opposite sides are congruent.
    So, it is a parallelogram.

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  2. http://img15.imageshack.us/img15/857/problem242a.png

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  3. See the drawing

    - ΔACB’ and ΔBCA’ are equilateral => ∠ACB’ =∠BCA’ = Π/3
    - ∠ACB = ∠ACB’ - ∠B’CB = Π/3 - ∠B’CB
    - ∠B’CA’ = ∠BCA’ - ∠B’CB = Π/3 - ∠B’CB
    - =>∠ACB =∠B’CA’
    - ΔACB is congruent to ΔB’CA’ (SAS) => BA=B’A’
    In the same way:
    - ΔCAB’ and ΔBAC’ are equilateral => ∠CAB’ =∠BAC’ = Π/3
    - ∠CAB = ∠CAB’ - ∠BAB’ = Π/3 - ∠BAB’
    - ∠B’AC’ = ∠BAC’ - ∠BAB’ = Π/3 - ∠BAB’
    - =>∠CAB =∠B’AC’

    - ΔCAB is congruent to ΔB’AC’ (SAS) => BC=B’C’
    - BA=B’A’ and BA=BC’ (ΔABC’ equilateral) => B’A’=BC’
    - BC=B’C’ and BC=BA’ (ΔBCA’ equilateral) => B’C’=BA’
    - => B’A’=BC’ and B’C’=BA’
    - => B’A’BC’ is a parallelogram

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  4. Tr. ABC is congruent to AB'C' and also to A'B'C, SAS in each case.

    So B'C' = BC = A'B and also A'B' = AB = BC'

    So the opposite sides of quadrilateral A'BC'B' are equal whence the same is a parallelogram

    Sumith Peiris
    Moratuwa
    Sri Lanka

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