Quadrilateral ABCD is convex. Let M be the midpoint of AD, and let N be the midpoint of BC. Prove that MN is greater than or equal to (AB+CD)/2.
If MN=(AB+CD)/2, then what can you say about quadrilateral ABCD.
Let P be the midpoint of diagonal BD. Then by the trangle inequality, MP + PN > MN. Also, BP + PN > BN and DP + PM > DM. Then 2DP + 2PN > BC and 2MP + DP > AD. Subtracting these inequalities, we get 2PN - 2MP > BC - AD. Putting this together with the inequality MP + PN > MN gives you MN < (AB + CD)/2.
If MN = (AB + CD)/2, then everything must be parallel, so ABCD is a rectangle.
