In figure 2 :-P and xq …
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Sia ? 4 years, 10 months ago
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
XP = XQ, ...(i) [tangents from X]
AP = AR, ... (ii) [tangents from A]
BR = BQ. ... (in) [tangents from B]
Now, XP = XQ {tex}\Rightarrow{/tex} XA + AP = XB + BQ
XA + AR = XB + BR [using (ii) and (iii)]
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