Question

A circle touches the sides of a quadrilatieral ABCD at P, Q, R, S respectively. The angles subtended at the centre by a pair of opposite sides have theirs sum as:

Answer 1

ABCD is a quadrilateral circumscribing the circle with a center O.
ABCD touches the circle at points, P,Q,R and S.
In $△AOP$ and $△AOS.$
$AP=AS$(Length of tangents draw from external point to a circle are equal)
$AO=AO($Common)
$OP=OS($both radius)
$\therefore △AOP\cong △AOS($SSS congruence rule)
$\angle AOP=\angle AOS\left(CPCT\right)$
Now lets number the angles for it to look easier.
Q from the above conclusion, we can prove.
$\angle 2=\angle 3⟶\left(1\right)\angle 4=\angle 5⟶\left(2\right)\angle 6=\angle 7⟶\left(3\right)$
Now,
$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360°$(sum of circle round a point is 360°)
$\angle 1+\angle 2+\angle 2+\angle 5+\angle 5+\angle 6+\angle 6+\angle 1=360°$(from $\left(1\right),\left(2\right),\&\left(3\right))$
$2(\angle 1+\angle 2+\angle 5+\angle 6)=360°(\angle 1+\angle 2)+(\angle 1+\angle 2)=180\angle AOB+\angle COD=180$
Thus, both the angles are supplementary.
similarly, we can prove
$\angle BOC+\angle AOD=180.$
Hence proved