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From the figure, there is 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,

AP=AR

BQ=BR

XP=XA+AP --------- (1)

XQ=XB+BQ --------- (2)

By substituting AP=AR in equation (1) and BQ=BR in equation (2), we get

XP=XA+AR

XQ=XB+BR

Since the tangents XP and XQ are equal, we get

XA+AR=XB+BR.

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