How to prove that any segment whose ends lie on the two bases of a trapezoid is bisected by the midsegment of the trapezoid

Looking at the image I have attached (with MN being the midsegment of the trapezoid), For any PQ with P and Q lying on the bases of the trapezoid there exists ON with O being the point of intersection between PQ and MN. ON must be parallel to PC because ON is part of MN and PC is a part of DC, and MN and DC are parallel. And N must be the midpoint of CB because MN is the midsegment of the trapezoid, so MN must bisect CB. And since ON is parallel to PC and bisects CB, that means it is a midsegment of QBCP and must bisect PQ.

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.