Urgent geometry help is needed here. Please help?

Question

ABCD is a trapezium with AB II CD. E and F are the midpoints of AD and BC respectively. Prove that EF II AB and EF = (AB + CD)/2.

Answer 1

A) First bring BD so that way you have two triangles ABD and BDC. Than you bring a parallel line to AB from E. Name the point that BD cuts the parallel K. Than:
For the triangle ABD: 1) E midpoint of AD
2) EK II AB
So K is the midpoint of BD and EK=AB/2
For the triangle BDC: 1) K midpoint of BD
2) KF II CD
So F is the midpoint of BC and KF=CD/2
And since the parallel from E passes through F you've proved EF II AB.
B) From what we proved before we have:
EK + KF=AB/2 + CD/2=(AB + CD)/2

Answer 2

Here's the general idea... you can work out the details.

1) drop the perpendiculars from A to CD and from B to CD (or the extension of CD). This will create some right triangles at both ends.

2) The segment joining E to the midpoint of the perpendicular creates another right triangle which is similar to the larger right triangle in a 1:2 ratio. Same with F at the other end.

3) From this you show that E, F and the midpoints of the perpendiculars are all parallel to AB and at the same height so they are collinear.

4) Add up the segments to show EF = (AB + CD)/2, using the 1:2 ratio again.

Give Molly (below) the 10 points though. She's not as messy as I am...

Answer 3

Draw he line AC intersecting EF at X.
X, E, and F are collinear because EX and XE share X.
Thus X is the midpoint of AC
Thus EX || DC and = 1/2 DC by the midline theorem.
Similarly, XF || AB and = 1/2 AB
Thus EX+XF = EF = 1/2DC+1/2AB = 1/2(AB+BC).
EF || AB || DC because EX ||DC and XF||AB.