Problem Statement
A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$.
Verified Solution & Marking Scheme
Label Points of Contact
Let the circle touch sides $AB, BC, CD, DA$ at points $P, Q, R, S$ respectively.
Apply Tangents from External Point Theorem
Lengths of tangents drawn from an external point to a circle are equal:
$AP = AS$ ...(1)
$BP = BQ$ ...(2)
$CR = CQ$ ...(3)
$DR = DS$ ...(4)
Add All Four Equations
$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$
Since $AP + BP = AB$, $CR + DR = CD$, $AS + DS = AD$, and $BQ + CQ = BC$:
$AB + CD = AD + BC$