0 is the point of intersection of the diagonals AC and BD

Solution:

Given ABCD is a trapezium. Diagonals AC and BD are intersect at 0.

PQ||AB||DC.    

To prove    $P O=Q O$

Proof in $\triangle A B D$ and $\triangle P O D$, $P O \| A B$ $[\because P Q \| A B]$

$\angle D=\angle D \quad$ [common angle]

$\angle A B D=\angle P O D \quad$ [corresponding angles]

$\therefore \quad \triangle A B D \sim \Delta P O D$ [by AAA similarity criterion]

Then, $\frac{O P}{A B}=\frac{P D}{A D}$ ...(i)

In $\triangle A B C$ and $\triangle O Q C, \quad O Q \| A B \quad[\because O Q \| A B]$

$\angle C=\angle C \quad$ [common angle]

$\angle B A C=\angle Q O C \quad$ [corresponding angle]

$\angle B A C=\angle Q O C$ [corresponding angle]

$\therefore$  $\triangle A B C \sim \triangle O Q C$ [by AAA similarity criterion]

Then, $\frac{O Q}{A B}=\frac{Q C}{B C}$ .....(ii)

Now, in $\triangle A D C$, $O P \| D C$

$\therefore$ $\frac{A P}{P P}=\frac{O A}{O C}$ [by basic proportionality theorem].....(iii)

In $\triangle A B C$. $O Q \| A B$

$\therefore$ $\frac{B Q}{Q C}=\frac{O A}{O C}$ [by basic proportionality theorem].....(iv)

From Eqs. (iii) and (iv), 

$\frac{A P}{P D}=\frac{B Q}{Q C}$

Adding 1 on both sides, we get

$\frac{A P}{P D}+1=\frac{B Q}{Q C}+1$

$\Rightarrow$ $\frac{A P+P D}{P D}=\frac{B Q+Q C}{Q C}$

$\Rightarrow$ $\frac{A D}{P D}=\frac{B C}{Q C}$

$\Rightarrow$ $\frac{O P}{A B}=\frac{O Q}{B C}$  [from Eqs. (i) and (ii)]

$\Rightarrow$ $\frac{O P}{A R}=\frac{O Q}{A R}$ [from Eq. (ii)]

$\Rightarrow$ $O P=O Q$ Hence proved.