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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