1. **Problem statement:** Given trapezium ABCD with E and F as midpoints of non-parallel sides AD and BC respectively, prove that line segment EF is parallel to AB and that its length is half the sum of lengths AB and CD. 2. **Key properties and formula:** In a trapezium, the segment joining the midpoints of the non-parallel sides is called the mid-segment. The mid-segment theorem states: $$EF \parallel AB \quad \text{and} \quad EF = \frac{AB + CD}{2}$$ 3. **Step-by-step proof:** - Let the coordinates of points be: $$A(x_1,y_1), B(x_2,y_2), C(x_3,y_3), D(x_4,y_4)$$ - Since E and F are midpoints: $$E = \left(\frac{x_1 + x_4}{2}, \frac{y_1 + y_4}{2}\right), \quad F = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)$$ - Vector EF is: $$\overrightarrow{EF} = F - E = \left(\frac{x_2 + x_3}{2} - \frac{x_1 + x_4}{2}, \frac{y_2 + y_3}{2} - \frac{y_1 + y_4}{2}\right) = \frac{1}{2} \left((x_2 + x_3) - (x_1 + x_4), (y_2 + y_3) - (y_1 + y_4)\right)$$ - Vector AB is: $$\overrightarrow{AB} = (x_2 - x_1, y_2 - y_1)$$ - Vector DC is: $$\overrightarrow{DC} = (x_3 - x_4, y_3 - y_4)$$ - Adding vectors AB and DC: $$\overrightarrow{AB} + \overrightarrow{DC} = (x_2 - x_1 + x_3 - x_4, y_2 - y_1 + y_3 - y_4) = ((x_2 + x_3) - (x_1 + x_4), (y_2 + y_3) - (y_1 + y_4))$$ - Therefore: $$\overrightarrow{EF} = \frac{1}{2} (\overrightarrow{AB} + \overrightarrow{DC})$$ 4. **Conclusion:** Since EF is a linear combination of AB and DC, and AB is parallel to DC (by trapezium definition), EF is parallel to AB. Also, the length of EF is half the sum of lengths AB and CD: $$EF = \frac{AB + CD}{2}$$ Hence, proved that: $$EF \parallel AB \quad \text{and} \quad EF = \frac{AB + CD}{2}$$