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


Answer:


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

    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^@.

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