Question

prove that the line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides and is equal to the half of their difference.

Answer 1

Prove that the line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides and is equal to the half of their difference.

Since, AB $\parallel$ DC and transversal AC cuts then at A and C respectively.

$$\therefore \angle 1 = \angle 2 \text{ (Alternate angles)}$$

Now, In $\Delta$APR and $\Delta$DPC,

$$\angle 1 = \angle 2$$

$\mathrm{AP} = \mathrm{CP}$ (P is mid point of AC)

$$\angle 3 = \angle 4 \text{ (Vertically opposite angles)}$$

$\therefore \Delta \mathrm{APR} \cong \Delta \mathrm{DPC}$

$$\Rightarrow \mathrm{AR} = \mathrm{DC} \text{ and } \mathrm{PR} = \mathrm{DP}$$

In $\Delta$DRB, P and Q are the mid-points of sides DR and DB respectively.

$\therefore \mathrm{PQ} \parallel \mathrm{RB}$

$$\Rightarrow \mathrm{PQ} \parallel \mathrm{AB}$$ $$\Rightarrow \mathrm{PQ} \parallel \mathrm{AB} \text{ and } \mathrm{DC}$$

Again P and Q are the mid-points of sides DR and DB respectively.

$$\therefore \mathrm{PQ} = \frac{1}{2} \mathrm{RB}$$ $$\Rightarrow \mathrm{PQ} = \frac{1}{2} (\mathrm{AB} - \mathrm{AR})$$ $$\Rightarrow \mathrm{PQ} = \frac{1}{2} (\mathrm{AB} - \mathrm{DC})$$