Question

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

Answer 1

name the point s that touches the circle as p,q,r,s
Ap=AS(tangents from point a)
BP=BS(tangents from point b)
CR=CQ(tangents from point c)
DR=DS(""""""'''""""""""""" point d)
adding.
AP+BP+ CR+ DR=AB+CQ+DS
therefore AB + CD =AD + BC

Answer 2

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According to the Question

Suppose the Quadrilateral ABCD drawn to Circumscribe a circle

The circle touches the side AB , BC , CD and DA

Hence

Length of 2 tangent drawn from an external point of circle are equal

AP = AS

BP = BQ

DR = DS

CR = CQ

Adding all the above we will get :-

(AP + BP)  + (CR + RD) = (BQ + QC) + (DS + SA)

AB + CD = BC + DA

PROVED

REASON = When you add AP and BP you will get AB and Similarly all above.