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A quadrilateral ABCD is drawn to circumscribe a circle,

Prove thatAB + CD = AD + BC

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by rohitpihwal7

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Prove thatAB + CD = AD + BC

by rohitpihwal7

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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!!!!!

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.