Question

In the adjoining figure, ABCD is a trapezium in which AB || DC. If M and N are the midpoints of BD and AC, respectively, prove that MN = `1/2` (CD – AB).

Answer 1

Given: ABCD is a trapezium with AB || CD. M and N are the midpoints of BD and AC respectively.

To Prove: MN = `1/2` (CD – AB).

Proof [Step-wise]:

1. Place the trapezium on a coordinate plane with the parallel sides horizontal.

Let AB have length a and lie on the line y = h with A = (x₁, h) and B = (x₁ + a, h).

Let CD have length c and lie on the line y = 0 with D = (x₂, 0) and C = (x₂ + c, 0).

This is possible because AB || CD.

2. Coordinates of midpoint N of AC:

`N = ((x_1 + (x_2 + c))/2, (h + 0)/2)` 

= `((x_1 + x_2 + c)/2, h/2)`

3. Coordinates of midpoint M of BD:

`M = (((x_1 + a) + x_2)/2, (h + 0)/2)` 

= `((x_1 + x_2 + a)/2, h/2)`

4. Compute the vector from M to N:

MN = N – M

= `((x_1 + x_2 + c)/2 - (x_1 + x_2 + a)/2, h/2 - h/2)` 

= `((c - a)/2, 0)`

5. The length of MN is the absolute value of the x-component:

`|(c - a)/2| = (c - a)/2`, taking CD ≥ AB so c ≥ a. 

Hence, MN = `1/2` (CD – AB).