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.
