Question

Abcd is quadrilateral circumscribing a circle then prove ab+cd=ad+bc

Answer 1

To Prove - AB + CD = AD + BC

Proof - Let AB touches the circle at P. BC touches the circle at Q. DC touches the circle at R. AD touches the circle at S.

THEN,PB=QB  ( Length of the the tangents drawn from the external point are always equal )

  QC =RC " 

  AP=AS  "

  DS=DP  "

NOW,  AB + CD

  = AP + PB+DR+RC

 = AS+QB+DS+CQ

  = AS+DS+QB+CQ

  = AD+BC

  HENCE PROVED

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Answer 2

because the lengths of tangents drawn from an external point to a circle are equal.

AP=AS

equation 1

BP=BQ

equation 2

CR=CQ

equation 3

DR=DS

equation 4

adding all the equations

AP+BP+CR+DR=AS+BQ+CQ+DS

AB+CD=AD+BC

hence prove