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  • In fig 2 a quadrilateral ABCD is drawn to circumscribe a circle, with center O, in such a way that the sides AB, BC, CD, and DA  touch the circle at the points P, Q, R, and S respectively. Prove that AB + CD = BC + DA 

In fig 2 a quadrilateral ABCD is drawn to circumscribe a circle, with center O, in such a way that the sides AB, BC, CD, and DA  touch the circle at the points P, Q, R, and S respectively. Prove that AB + CD = BC + DA 

 

 

 

 

 

 
 
 
 
 

Answers (1)

Given ABCD is quadrilateral  P,Q,R,S are point of contact 

To prove  AB + CD = BC + DA 

 theorem -

tangents drawn from an external point are equal 

Using the above theorem we can say 

AP = AS --- (1) 

BP = BQ --- (2) 

CR = CQ --- (3) 

OQ = OS --- (4) 

Adding (1) , (2) , (3) , (4) 

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

 

Posted by

Ravindra Pindel

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