Let's prove that the midpoints of the sides of any quadrilateral form a parallelogram. Since trapezoids are quadrilaterals, this shows the claim we are asked to prove. Usually, placing a quadrilateral in a special position of the coordinate plane helps the calculation, but in this case we can use a general diagram.
We can use the Midpoint Formula to find the coordinates of the midpoints of the sides.
Endpoints
Midpoint
A(a_x,a_y) and B(b_x,b_y)
E(a_x+b_x/2,a_y+b_y/2)
B(b_x,b_y) and C(c_x,c_y)
F(b_x+c_x/2,b_y+c_y/2)
C(c_x,c_y) and D(d_x,d_y)
G(c_x+d_x/2,c_y+d_y/2)
D(d_x,d_y) and A(a_x,a_y)
H(d_x+a_x/2,d_y+a_y/2)
Let's add these coordinates to the diagram.
Remember, we would like to show that EFGH is a parallelogram. We have several possible ways to continue.
We can show that opposite sides are parallel, so by definition we have a parallelogram. We can use the Slope Formula to follow this approach.
We can show that opposite sides are congruent, so by Theorem 6-8 we have a parallelogram. We can use the Distance Formula to follow this approach.
We can show that the diagonals bisect each other, so by Theorem 6-11 we have a parallelogram. We can use the Midpoint Formula to follow this approach.
Since we already used the Midpoint Formula, let's do the calculations for the third approach. We will find the midpoint of diagonal EG first.
We can see that the midpoint of EG is the same as the midpoint of FH. This means that the diagonals EG and FH bisect each other, so according to Theorem 6-11 quadrilateral EFGH is a parallelogram.
