A quadrilateral `ABCD` is drawn to circumscribe a circle, as shown in the figure. Prove that `AB+CD=AD+BC`.

Explanation

We know that the lengths of tangents drawn from an exterior point to a circle are equal.
`:.AP=AS`, ……..`(i)` [tangents from `A`]
`BP=BQ`, ……..`(ii)` [tangents from `B`]
`CR=CQ`, …….`(iii)` [tangents from `C`]
`DR=DS`, ……….`(iv)` [tangents from `D`]
`:.AB+CD=(AP+BP)+(CR+DR)`
`=(AS+BQ)+(CQ+DS)` [using `(i)`, `(ii)`, `(iii)`, `(iv)`]
`=(AS+DS)+(BQ+CQ)`
`=AD+BC`.
Hence, `AB+CD=AD+BC`.