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.
Answer:
- We know that the lengths of tangents drawn from an external point to a circle are equal.
Thus [Math Processing Error] - We also see that
XP=XA+APXP=XA+AP
and
XQ=XB+BQXQ=XB+BQ.
Thus, [Math Processing Error] - Thus, XA+AR=XB+BRXA+AR=XB+BR.