### In the given figure, $XP$ and $XQ$ are the two tangents to a circle with center $O$ drawn from an external point $X$. $AB$ is another tangent touching the circle at $R$. Prove that $XA + AR = XB + BR$. O X A B P R Q

Thus \begin{aligned} & XP = XQ && \text{[Tangents from X]} && \ldots \text{(i)} \\ & AP = AR && \text{[Tangents from A]} && \ldots \text{(ii)} \\ & BR = BQ && \text{[Tangents from B]} && \ldots \text{(iii)} \end{aligned}
2. We also see that $XP = XA + AP$ and $XQ = XB + BQ$.
Thus, \begin{aligned} & XP = XQ && \text{[Using (i)]} \\ \implies & XA + AP = XB + BQ \\ \implies & XA + AR = XB + BR && \text{[Using } eq \text{ (ii) and } eq \text{ (iii)]} \end{aligned}
3. Thus, $XA + AR = XB + BR$.