A quadrilateral A B C D is drawn to circumscribe a circle.Prove that, A B + C D = A D + B C

Given that, A B C D is quadrilateral. To Prove: A B + C D = A D + B C Now, let A B touches the circle at P , B C at Q , D C at R and A D at S . Thus PB = QB (length of tangents drawn from an external point are always equal) QC = RC AP = AS DS = DP Now, A B + C D = A P + P B + D R + R C = A S + Q B + D S + C Q = A S + D S + Q B + C Q = A D + B C Hence, A B + C D = A D + B C [Proved]