Question

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB+CD=AD+BC

Answer 1

Step-by-step explanation:to prove:

AB+CD=AD+BC

Proof :

AP=AS

PB=BQ

CQ=CR

RD=DS (the length of tangent drawn from an external point to a circle are equal)

AP+PB+RD+CR=AS+BQ+DS+CQ

=>AB+CD=AD+BC

Hence proved

Answer 2

Here is your answer mate ↙️↙️↙️✅✅✅

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

Hope this helps you dude ☺️☺️✌️✌️✌️