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Question
In Figure 2, XP and XQ are two tangents to the circle with centre O, drawn from an external point X. ARB is another tangent, touching the circle at R. Prove that XA + AR = XB + BR ?
Solution
In the given figure, we have an external point X from where two tangents, XP and XQ, are drawn to the circle.
XP = XQ (The lengths of the tangents drawn from an external point to the circle are equal.)
Similarly, we have:
AP = AR
BQ = BR
Now, XP = XA + AP ...(1)
XQ = XB + BQ ...(2)
On putting AP = AR in equation (1) and BQ = BR in equation (2), we get:
XP = XA + AR
XQ = XB + BR
Since XP and XQ are equal, we have:
XA + AR = XB + BR
Hence, proved.
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