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.

## Monday, February 2, 2009

### Elearn Geometry Problem 240: Triangle with Equilateral triangles, Parallelogram

Labels:
60,
congruence,
equilateral,
parallelogram,
triangle

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ang C'AC = 60 - A =>

ReplyDeleteB'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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1. Triangles AA’B’ and ABC are congruent ( AB’=AC , AC’=AB, angle B’AC’= angle CAB ) so BC’=BC=BA’

ReplyDelete2. 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.