2. Draw a figure to illustrate part of the infinite pattern that can be derived from a given equilateral triangle ABC by applying all the vectors con- sisting of an integral multiple of AB plus an integral multiple of AC.
2. Draw a figure to illustrate part of the infinite pattern that can be derived from a given equilateral triangle ABC by applying all the vectors con- sisting of an integral multiple of AB plus an integral multiple of AC.
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Answer just #2 please proof

Transcribed Image Text:9:45 1
LTE
A aproged.pt
positions of the circle meet at B and C, these are two vertices of the
desired rectangle ABCD, whose sides AB and DC are equal and
parallel to a.
82
TRANSFORMATIONS
Figure 4.1D
EXERCISES
1. In AABC (Figure 4.1E) “inscribe" a line segment equal and parallel
to the given segment a.
2. Draw a figure to illustrate part of the infinite pattern that can be derived
from a given equilateral triangle ABC by applying all the vectors con-
sisting of an integral multiple of AB plus an integral multiple of AC.
Figure 4.1E
4.2 Rotation
Another kind of transformation that preserves distance is rotation.
Here the entire plane is turned about some point through a given angle.
Thus the size and shape of any figure are kept invariant, but its points
all move along arcs of concentric circles. The center (which may or
may not “belong" to the figure being rotated) is the only point that
remains fixed.
As an example of the use of a rotation, let us consider AABC (Figure
4.2A) with equilateral triangles BPC, CQA, ARB erected (externally)
on the three sides. After drawing the lines BQ and Cr, which meet at
F, we observe that a rotation through 60° about A takes AARC into
AA BQ. Hence ZRFB = 60° and RC = BQ. Similar reasoning
shows that PA = CR. Thus
AP = BQ = CR.
ROTATION
83
Moreover, since
ZRFB = 60° = ZRAB and 2CFQ = 60° = 2CAQ,
the quadrangles A RBF and CQAF are cyclic; and since ZBFC = 120°
while 2CPB = 60 PEPi
drangle. Therefore the
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