Question

If a quadrilateral ABCD is circumscribed by a circle, then prove that AB+CD=BC+DA

Answer 1

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Here is your solution
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✨To Prove✨
AB + CD= BC +DA

$\color{blue}{Proof}$
We know that tangents drawn from same external points are equal. So,
DR = DS [ Tangents drawn from point D]..(1)
CR=CQ [Tangents drawn from point C]..(2)
BP= BQ [Tangents drawn from point B]...(3)
AP=AS [Tangents drawn from point A]....(4)

On adding equation
1,2,3 and 4

DR+ CR +BP+ AP = DS +CQ +BQ +AS
(DR+CR)(BP+AP)= (DS+AS)(CQ+BQ)
CD + AB = AD + BC

$Proved$

⭐Hope it helps you⭐
➡️ Thank you⬅️

Answer 2

Step-by-step explanation:

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