Question

A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the given figure.

Prove that : AB + CD = AD + BC.

Answer 1

We know that lengths of tangents to circle from a fixed external point are equal. 

∴ AP = AS … (1) 

(∵ AP & AS are tangents to circle from point A) 

BP = BQ … (2) 

(∵ BP & BQ are tangents to circle from point B) 

CR = CQ … (3) 

(∵ CQ & CR are tangents to circle from point C) 

& DR = DS … (4) 

(∵ DR & DS are tangents to circle from point D) 

Now, 

Adding equations (1), (2), (3) and (4), we get 

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

⇒ AB + CD = (AS + DS) + BC

( ∵ AP + BP = AB, BQ + CQ = BC And CR + DR = CD) 

⇒ AB + CD = AD + BC. 

(∵ AS + DS = AD) 

Hence Proved.