what would this ratio be equal to in a trapezium?

Question

What would be the logical reason for the answer and could you mention the theorems used?

Answer 1

The answer is (2).

Draw a line, $L$, through $O$ parallel to the trapezoid's bases. Denote by $A$ $L$'s intersection with $\overline{PS}$, and denote by $B$ $L$'s intersection with $\overline{QR}$. Since a parallel to one of the sides of a triangle divides the other two sides proportionally,

$$ \frac{PO}{OR}\overset{\Delta PQR}{=} \frac{QB}{BR} \overset{\Delta SQR}{=} \frac{QO}{OS}. $$

Therefore $$ \begin{align} \frac{PO - OR}{OR} &= \frac{QO - OS}{OS}, \\ \frac{PO + OR}{OR} &= \frac{QO + OS}{OS}. \end{align} $$

Therefore

$$ \frac{PO-OR}{QO-OS} = \frac{OR}{OS} = \frac{PO+OR}{QO+OS} =\frac{PR}{QS}. $$