In the given figure, XPXPXP and XQXQXQ are the two tangents to a circle with center OOO drawn from an external point XXX. ABABAB is another tangent touching the circle at RRR. Prove that XA+AR=XB+BRXA+AR=XB+BRXA+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 [Math Processing Error]
  2. We also see that XP=XA+APXP=XA+AP and XQ=XB+BQXQ=XB+BQ.

    Thus, [Math Processing Error]
  3. Thus, XA+AR=XB+BRXA+AR=XB+BR.

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