In figure, XP and XQ are two tangents to a circle with center O from a point X outside the circle. ARB is tangent to circle at R. Prove that XA + AR = XB + BR.

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Solution

As the lengths of tangents drawn from an external point to a circle are equal.

XP = XQ

XA + AP = XB + BQ …..(i)

Also BQ = BR and AR = AP

Substituting in (i)

XA + AR = XB + BR

Hence proved.

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In figure, XP and XQ are two tangents to a circle with center O from a point X outside the circle. ARB is tangent to circle at R. Prove that XA + AR = XB + BR.

As the lengths of tangents drawn from an external point to a circle are equal. XP = XQ XA + AP = XB + BQ …..(i) Also BQ = BR and AR = AP Substituting in (i) XA + AR = XB + BR Hence proved.

As the lengths of tangents drawn from an external point to a circle are equal.

XP = XQ

XA + AP = XB + BQ …..(i)

Also BQ = BR and AR = AP

Substituting in (i)

XA + AR = XB + BR

Hence proved.