ABCD is a trapezium and P, Q are the mid points of the diagonals AC and BD respectively. Then PQ is equal to

In a trapezium, the line segment connecting the midpoints of the diagonals is equal to half the sum of the lengths of the bases.

Here are the steps to prove it:

1. Let's denote the bases of the trapezium ABCD as AD and BC, and the diagonals as AC and BD. Let's denote the midpoints of AC and BD as P and Q respectively.

2. Since P and Q are midpoints, AP = PC and BQ = QD.

3. Now, consider triangles APB and CPD. They are similar because they have two pairs of congruent sides (AP = PC and BP = PD) and the included angles are equal (angle APB = angle CPD).

4. Because the triangles are similar, the ratio of corresponding sides is equal. Therefore, PQ/AB = PD/CD.

5. Solving for PQ gives PQ = (AB + CD)/2, which is half the sum of the lengths of the bases.

So, the line segment PQ is equal to half the sum of the lengths of the bases of the trapezium ABCD.