### A quadrilateral $ABCD$ is drawn to circumscribe a circle, as shown in the figure. Prove that $AB + CD = AD + BC$. D C A B R S Q P

Step by Step Explanation:
1. We know that the lengths of tangents drawn from an exterior point to a circle are equal.

Thus, \begin{aligned} & AP = AS && \ldots \text{(i)} && \text{[Tangents from A]} \\ & BP = BQ && \ldots \text{(ii)} && \text{[Tangents from B]} \\ & CR = CQ && \ldots \text{(iii)} && \text{[Tangents from C]} \\ & DR = DS && \ldots \text{(iv)} && \text{[Tangents from D]} \end{aligned}
2. Now, let us find the length $AB + CD$.

We see that $AB = AP + BP \text{ and } CD = CR + DR.$

So, \begin{aligned} AB + CD & = (AP + BP) + (CR + DR) \\ & = (AS + BQ) + (CQ + DS) && \text{[Using } eq \text{(i), (ii), (iii), and (iv)]} \\ & = (AS + DS) + (BQ + CQ) \\ & = AD + BC && [ \because \text{ AD = AS + DS and BC = BQ +CQ ]} \end{aligned}
3. Hence, $AB + CD = AD + BC$.