See complete Problem 243 at:

gogeometry.com/problem/p243_triangle_equilateral_parallelograms.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, February 4, 2009

### Triangle with Equilateral triangles, Rhombuses, Parallelograms

Labels:
congruence,
equilateral,
parallelogram,
rhombus,
triangle

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a) A'BA"C has

ReplyDeleteA"C//BA', A"B//CA' ( ang 120°+60°)

BC perpendicular to A"A' (A"A' altitude for equilat tr)

=> A'BA"C is rhombus

b) tr ABC "goes" to B"CA', AB"C', A"CB' (see P240)

A'BC'B" has

ang C'B"A' = 360°- (C+60+A)=300+B-(A+B+C)

ang C'B"A' = 120+B

ang C'BA' = 120+B

=>

C'B"A' = C'BA' (1)

ang BC'B" = 60 - B

ang BA'B" = 60 - B

=>

BC'B" = BA'B" (2)

(1) & (2) =>

BC'B"A' paralelogram

c) A"C'B"C has

ang B"C'A" = 60 - (60 - B) - (60 - A)

ang B"C'A" = A + B - 60

ang A"CB" = C - (C - 60) - (C - 60) = 120 - C

= 120 - (180 - B - A)= A + B - 60

B"C'A" = A"CB" (1)

ang C'B"C = C + 60

ang C'A"C' = C + 60

C'B"C = C'A"C' (2)

(1) & (2) =>

C'A"CB" paralelogram

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