$A$ quadrilateral $ABCD$ is drawn to circumscribe a circle (see figure). Prove that $AB + CD = AD + BC$.
(N/A) We know that the lengths of tangents drawn from an external point to a circle are equal.
From point $D$,$DR = DS$ ... $(1)$
From point $C$,$CR = CQ$ ... $(2)$
From point $B$,$BP = BQ$ ... $(3)$
From point $A$,$AP = AS$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
Rearranging the terms,we get:
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
Since $DR + CR = CD$,$BP + AP = AB$,$DS + AS = AD$,and $CQ + BQ = BC$,we have:
$CD + AB = AD + BC$
Hence,$AB + CD = AD + BC$.
From point $D$,$DR = DS$ ... $(1)$
From point $C$,$CR = CQ$ ... $(2)$
From point $B$,$BP = BQ$ ... $(3)$
From point $A$,$AP = AS$ ... $(4)$
Adding equations $(1), (2), (3),$ and $(4)$,we get:
$DR + CR + BP + AP = DS + CQ + BQ + AS$
Rearranging the terms,we get:
$(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)$
Since $DR + CR = CD$,$BP + AP = AB$,$DS + AS = AD$,and $CQ + BQ = BC$,we have:
$CD + AB = AD + BC$
Hence,$AB + CD = AD + BC$.
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