Question

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

Answer 1

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 [From (i), (ii), (iii) and (iv)]
= (AS + DS) + (BQ + CQ)
= AD + BC
Thus, AB + CD = AD + BC