Question

A quadrilateral ABCD is drawn to circumscribe a circle,
Prove thatAB + CD = AD + BC

Answer 1

DR=DS [TANGENTS FROM EXTERNAL POINT ARE EQUAL]---------(1)
similarily,
CR=CQ-----------------------(2)
BP=BQ-----------------------(3)
AP=AS-----------------------(4)
adding all the above equations,we get
DR+CR+BP+AP=DS+CQ+BQ+AS
(DR+CR)+(BP+AP)=(DS+AS)+(CQ+BQ)
CD+AB=AD+BC
hence proved!!!!!

Answer 2

Proof:
Let the sides AB, BC, CD and AD meet the circle at P, Q, R and S.
Therefore, A is an external point from which 2 tangents AP and AS are drawn.
                 ⇒ AP=AS    --------------(1) 
Similarly ,      BP=BQ    --------------(2)
                    CR=CQ    ---------------(3)
                    DR=DS    ---------------(4)

Adding eq.s (1), (2), (3),(4) ,we get 
  AP+BP+CR+DR=AS+DS+BQ+CQ
          ⇒AB + CD=AD + BC

Hence proved.