If the side of quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD

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Solution

Let a circle touch the sides $A B, B C, C D,$ and $D A$ of a quadrilateral $A B C D$ at $P, Q, R$ $a n d$ S $r e s p e c t i v e l y A s,$ AP $a n d$ AS $a r e t a n g e n t s t o t h e c i r c l e f r o m a n e x t e r n a l p o i n t$ A $, w e h a v e$ AP = AS .... ( 1) $S i m i l a r l y, w e a l s o g e t$ BP = BQ ..... (2)CR = CQ .....(3) DR = DS.....(4) $A d d i n g (1), (2), (3) a n d (4), w e g e t$ AP + BP +CR + DR = AS +DS +BQ +CQ AB +CD = AD +BC $T h e r e f o r e,$ AB + CD = AD + BC$
-Hence proved

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