Monday, February 2, 2009

Elearn Geometry Problem 240: Triangle with Equilateral triangles, Parallelogram

Problem: Triangle with Equilateral triangles, Parallelogram

See complete Problem 240 at:
gogeometry.com/problem/p240_triangle_equilateral_parallelogram.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. ang C'AC = 60 - A =>
    B'AC' = A
    A'CB' = C

    extend BA' to P ( P on AB'), BC' to G, ( G on B'C )
    BPB' = B - 60 + 60 + A, BGB' = B - 60 + 60 + C =>

    BPB' = A + B
    BGB' = B + C

    ▲ABC and ▲AC'B' are congruent
    1) AB = AC'
    2) BAC = C'AB' = A
    3) AC = AB'
    =>
    AC'B' = B, AB'C' = C
    =>
    BPB' + AB'C' = (A + B) + C = 180°
    =>

    BP//B'C'

    BGB' + GB'A' = (B + C) + A = 180°
    =>

    BC'//A'B'
    ------------------------------------------

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  2. 1. Triangles AA’B’ and ABC are congruent ( AB’=AC , AC’=AB, angle B’AC’= angle CAB ) so BC’=BC=BA’
    2. Triangles CA’B’ and CBA are congruent ( CB’=CA, CA’=CB, angle B’CA’= angle ACB)
    so B’A’=BA=BC’
    3. Quadrilateral BA’B’C’ have opposite sides congruent so BA’B’C’ is a parallelogram.

    ReplyDelete