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.

4 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.

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  3. See the drawing

    ΔCAB’ and ΔC’AB are equilateral
    => ∠CAB’=∠C’AB=Π/3
    ∠BAB’=a+∠CAB’=∠BAC’+a’
    => a=a’
    AC’=AB and AB’=AC and a=a’=> ΔBAC and ΔB’AC’ are congruent (SAS)
    =>B’C’=BC
    But BC=BA’=>B’C’=BA’
    In the same way : ∠BCA’=∠B’CA=Π/3
    ∠BCA=c= Π/3 - ∠ACA’
    ∠B’CA’=c’= Π/3 - ∠ACA’ =>c=c’
    CB=CA’ and CA=CB’ and c=c’=> ΔBCA and ΔA’CB’ are congruent (SAS)
    =>AB=A’B’
    But AB=BC’ => A’B’=BC’

    B’C’=BA’ and A’B’=BC’ => BA’B’C’ is a parallelogram



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  4. It is not difficult to see that < A'CB' = C and hence Tr.s A'B'C & ABC are congruent whence A'B' = BC'

    By similar reasoning A'B = B'C'

    In quadrilateral A'B'C'B the opposite sides are thus equal and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete