If ABCD is a cyclic quadrilateral inscribed in a circle with center O, then prove that: \[ AB + CD = BC + DA \]

Q.If ABCD is a cyclic quadrilateral inscribed in a circle with center O, then prove that: \[ AB + CD = BC + DA \]

Given: ABCD is a quadrilateral inscribed in a circle with center O. Let AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively. To Prove: \(AB + CD = BC + DA\) **Proof:** From external point A, two tangents are drawn to the circle: AP and AS. ∴ \(AP = AS\) Similarly, From point B: \(BP = BQ\) From point C: \(CQ = CR\) From point D: \(DR = DS\) Now, \[ AB + CD = AP + BP + CR + DR = AS + BQ + CQ + DS \] Group the terms: \[ = (AS + DS) + (BQ + CQ) = DA + BC \] \[ AB + CD = BC + DA \quad \text{(Proved)} \]