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Question

A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA                                   


Solution

Given, a quad. ABCD and a circle touches its all four sides at P,Q, R, and S respectively.



To prove: AB + CD = BC + DA
Proof:
          We know that tangents drawn from external point to a circle are equal.
So, AP=AS,
      BP=BQ,
      CQ=CR,
      DR=DS.

L.H.S. = AB + CD

= AP + PB + CR + RD

= AS + BQ + CQ + DS

(Tangents from same external point are always equal)

= (AS + SD) + (BQ + QC)

= AD + BC

= R.H.S.                                   
 Hence Proved.

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