Question

plz answer this question
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Answer 1

Given, ABCD is a quadrilateral.

So, AP = AS [ Tangents are equal ]

PB = BQ

CQ = CR

DR = DS

L. H. S. = AB + CD

= ( AP+ PB ) + ( CR + RD )

= ( AS + BQ ) + ( CQ + DS )

= ( AS + DS) + ( BQ + CQ )

= AD + BC

= R. H. S.

Hence, Proved.

Answer 2

given= let abcd is a quadrilateral circumscribed the circle at centre o.

the quadrilateral touches the circle at points p,q,r and s.

To prove=AB+CD=AD+BC

proof= from theorem 10.2 tangents drawn from external points to a circle.

AP=AS----1

BP=BQ-----2

CR=CQ-----3

DR=DS-------4

adding 1+2+3+4 we get

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

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

AB+CD=AD+BC.

HENCE PROVED.

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