In the given figure, `XP` and `XQ` are two tangents to the circle with centre `O`, drawn from an external point `X`. `ARB` is another tangent, touching the circle at `R`. Prove that `XA+AR=XB+BR`.

Explanation

We know that the lengths of tangents drawn from a exterior point to a circle are equal.
`:.XP=XQ`, ………`(i)`[tangents from `X`]
`AP=AR`, ………`(ii)` [tangents from `A`]
`BR=BQ`. ………..`(iii)` [tangents from `B`]
Now, `XP=XQimpliesXA+AP=XB+BQ`
`impliesXA+AR=XB+BR` [using `(ii)` and `(iii)`].