Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 10 - Section 10.4 - Analytic Proofs - Exercises - Page 465: 29

Answer

Yes, the shape would still be a parallelogram.

Work Step by Step

We can consider exercise 7 for any values of a,b,c,d, and e. $A: 0,0 \\ B: 2a, 2b \\ C: 2c, 2d \\ D: 2e, 0$ This means that the midpoints are: $(a,b); (a+c, b+d); (e,0); (e+c, d)$ Thus, the slopes are: $m_1 = \frac{b+d-b}{a+c-c} =\frac{d}{c}$ $m_2 = \frac{b+d-d}{a+c-e-c} =\frac{b}{a-e}$ $m_3 = \frac{d-0}{e+c-e} =\frac{d}{c}$ $m_4 = \frac{b-0}{a-e} =\frac{b}{a-e}$ We see that the shape is a parallelogram, for corresponding slopes are equal.
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