Prove that the median of a trapezoid bisects each diagonal of the trapezoid.?

Answer 1

Use basic proportionality theorem.
Let AB || CD || EF and AD be the diagonal meeting EF at G
Thus,
EG/ CD = AG/AD = AE/AC = 1/2
Thus AG : AD = 1/2 = > AG : GD = 1 : 1 =>the median bisects the diagonal.

Answer 2

I have no idea how to explain this online, but here's my shot:

1) Draw a trapezoid and label the bases parallel
2) Draw in the midsegment, and mark the parts of the bisected sides congruent.
3) Draw in one diagonal
4) Mark the midsegment parallel to the bases (this is a property you already know, right?)
5) Now look at one of the two big triangles formed by the diagonal, a base, and one of the sides. You should see that the side of the triangle (the base of the trapezoid) is parallel to a line going across on the inside of the triangle. You should also notice that this line bisects the other side of the triangle (the bisected side of the trapezoid).
6) This line that's cutting across the triangle is the midsegment of the triangle (you can prove this a number of ways - using a similarity argument would do the trick)
7) Because that line is the midsegment, the diagonal that represents the third side of the triangle must be bisected by that line.

The same applies to the same diagonal.