A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Then

(a): Using theorem, the lengths of tangents drawn from an external point to a circle are equal. A is an external point, then AP = AS  .... (i) B is an external point, then BP = BQ .... (ii) C is an external point, then CQ = RC .... (iii) D is an external point, then SD = RD .... (iv) On adding Eq. (i), (ii), (iii) and (iv), we get $\left(AP+BP\right)+\left(RC+RD\right)$ $=\left(AS+BQ\right)+\left(CQ+SD\right)$ $\Rightarrow$AB+CD=(AS+SD)+(BQ+CQ)⇒\displaystyle{A}{B}+{C}{D}={A}{D}+{B}{C}