Question

a circle touches all the four sides of a quadrilateral ABCD,then prove that AB+CD=BC+DA

Answer 1

Given : A circle with center O touching the sides of Quadrilateral ABCD .

To Prove - AB+CD=BC +DA

Construction- Mark the points touching the Quadrilateral as P,Q,R and S .

Proof - see the pic ☝️

ThAnKs !✌️✌️

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Answer 2

Answer:

Step-by-step explanation:

Given : ABCD is a quadrilateral .

To Prove : AB+CD=BC+DA

Solution :

Tangents from  the same point touches to circle are equal

Then ,

AP = AS [ tangent from point " A " ]

BP = BQ [ tangent from point " B " ]

DR = DS [ tangent from point " D " ]

CR = CQ [ tangent from point " C " ]

Add equations

AP + BP + DR + CR = AS + BQ + SD + QC

(AP + PB) + (DR + RC) = (AS + SD) + (BQ + QC)

$\bold{\boxed{\boxed{AB \: + \: DC \: = \: AD \: +  \: BC}}}$

Hence Proved

Thanks

For any clearance see the answer in link given below :-

https://brainly.in/question/1558912