Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.