Brazil
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 ^@.
Answer:
- 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} @^ - 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} @^ - Thus, ^@XA + AR = XB + BR^@.
