Question

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC. [ 2 Marks]

Answer 1

From the figure we observe that,
DR = DS (Tangents on the circle from point D) … (i)
AP = AS (Tangents on the circle from point A) … (ii)
BP = BQ (Tangents on the circle from point B) … (iii)
CR = CQ (Tangents on the circle from point C) … (iv)
[ 1 Mark ]
Adding all these equations,
DR + AP + BP + CR = DS + AS + BQ + CQ
$\Rightarrow$ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)
$\Rightarrow$ CD + AB = AD + BC
[ 1 Mark]