Question 11 Mark
A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$


Answer
View full question & answer→We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$AP = AS, ... (i)$ [tangents from $A$]
$BP = BQ, ... (ii)$ [tangents from $B$]
$CR = CQ, ... (iii)$ [tangents from $C$]
$DR = DS. ... (iv)$ [tangents from $D$]
$AB + CD = (AP + BP) + (CR + DR)$
$= (AS + BQ) + (CQ + DS) [using (i), (ii), (iii), (iv)]$
$= (AS + DS) + (BQ + CQ)$
$= AD + BC.$
Hence, $AB + CD = AD + BC.$
$AP = AS, ... (i)$ [tangents from $A$]
$BP = BQ, ... (ii)$ [tangents from $B$]
$CR = CQ, ... (iii)$ [tangents from $C$]
$DR = DS. ... (iv)$ [tangents from $D$]
$AB + CD = (AP + BP) + (CR + DR)$
$= (AS + BQ) + (CQ + DS) [using (i), (ii), (iii), (iv)]$
$= (AS + DS) + (BQ + CQ)$
$= AD + BC.$
Hence, $AB + CD = AD + BC.$



