Answer is A. AC + AD = BD + CD
Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
By the above property,
AP = AS (tangent from A)
BP = BQ (tangent from B)
CR = CQ (tangent from C)
DR = DS (tangent from D)
Now we add above 4 equations,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
[∵ AP + BP = AB
CR + DR = CD
AS + DS = AD
BQ + CQ = BC]
Hence, the right option is AB + CD = AD + BC