Question

Question 13
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer 1

Let ABCD be a quadrilateral circumscribing a circle with O such that it touches the circle at point P, Q, R, S. Join the vertices of the quadrilateral ABCD to the center of the circle.
In $\Delta$ OAP and $\Delta$ OAS,
AP = AS (Tangents from the same point)
OP = OS (Radii of the circle)
OA = OA (Common side)
$\Delta$ OAP $\cong$ $\Delta$ OAS (SSS congruence condition)
$\therefore \angle POA=\angle AOS$
$\Rightarrow \angle 1=\angle 8$
Similarly we get,
$\angle 2=\angle 3$
$\angle 4=\angle 5$
$\angle 6=\angle 7$
Adding all these angles,
$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8={360}^{\circ }$
$\Rightarrow (\angle 1+\angle 8)+(\angle 2+\angle 3)+(\angle 4+\angle 5)+(\angle 6+\angle 7)={360}^{\circ }$
$\Rightarrow 2\angle 1+2\angle 2+2\angle 5+2\angle 6={360}^{\circ }$
$\Rightarrow 2(\angle 1+\angle 2)+2(\angle 5+\angle 6)={360}^{\circ }$
$\Rightarrow (\angle 1+\angle 2)+(\angle 5+\angle 6)={180}^{\circ }\Rightarrow \angle AOB+\angle COD={180}^{\circ }$
Similarly, we can prove that $\angle BOC+\angle DOA={180}^{\circ }$
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.