8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that $AB + CD = AD + BC$
To prove- $AB + CD = AD + BC$
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
$AP =AS$.......(i)
$BP = BQ$.........(ii)
$AS = AP$...........(iii)
$CR = CQ$ ...........(iv)
By adding all the equations, we get;
$AP + BP +RD+ CR = AS +DS +BQ +CQ$
$\implies(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)$
$\implies AB + CD = AD + BC$
Hence proved.