A quadrilateral \(ABCD\) is drawn to circumscribe a circle (see Fig.). Prove that \(AB + CD = AD + BC\).
Proof:
We know that, The lengths of tangents drawn from an external point to a circle are equal.
Therefore, \(AP = \) ----(1)
\(BP = BQ\) ----(2)
\(CR = CQ\) -----(3)
\(DR = DS\) -----(4)
By, adding (1), (2), (3) and (4) \(RHS = LHS\) and simplifying then we get,
\(AP + BP + CR + DR= (AS + \) \()\)\(+ (BQ + \)\()\)
\(AB + CD = AD + BC\)
Hence, proved.