Answer
The midpoints of a trapezoid are connected to form a rhombus.
Work Step by Step
We call the midpoints (b,c), (a,2c), (2a-b,c), and (a,0). Thus, we use the distance formula:
$d_1 = \sqrt{(2c-c)^2 + (b-a)^2 } = \sqrt{c^2 + (b-a)^2}$
$d_2 = \sqrt{(2c-c)^2 + (a-2a+b)^2 } = \sqrt{c^2 + (b-a)^2}$
$d_3 = \sqrt{(c-0)^2 + (2a-b-a)^2 } = \sqrt{c^2 + (b-a)^2}$
$d_4 = \sqrt{(c-0)^2 + (b-a)^2 } = \sqrt{c^2 + (b-a)^2}$
The sides all have equal lengths, so the shape is a rhombus.