Q13:

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Given: A quadrilateral $ABCD$ circumscribing a circle with center $O$. The sides $AB, BC, CD,$ and $DA$ touch the circle at points $P, Q, R,$ and $S$ respectively.

To Prove: The opposite sides subtend supplementary angles at the center $O$. That is, $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.

Step 1: Joining the points of contact to the center.
Join the center $O$ to the points of contact $P, Q, R,$ and $S$. Also, join $OA, OB, OC,$ and $OD$.

Step 2: Identifying congruent triangles.
Consider $\triangle OAP$ and $\triangle OAS$:

• $AP = AS$ [Tangents drawn from an external point to a circle are equal in length]
• $OA = OA$ [Common side]
• $OP = OS$ [Radii of the same circle]

Step 3: Applying the same logic to other pairs.
Similarly, we can prove:

• $\triangle OBP \cong \triangle OBQ \implies \angle 3 = \angle 4$
• $\triangle OCQ \cong \triangle OCR \implies \angle 5 = \angle 6$
• $\triangle ODR \cong \triangle ODS \implies \angle 7 = \angle 8$

Step 4: Summing the angles around the center.
The sum of all angles around the center $O$ is $360^\circ$: $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$

Step 5: Substituting the equal angles.
Since $\angle 1 = \angle 2$, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, and $\angle 7 = \angle 8$, we substitute these into the equation: $2\angle 2 + 2\angle 3 + 2\angle 6 + 2\angle 7 = 360^\circ$ Dividing by 2: $\angle 2 + \angle 3 + \angle 6 + \angle 7 = 180^\circ$

Step 6: Grouping the angles.
Rearranging the terms: $(\angle 2 + \angle 3) + (\angle 6 + \angle 7) = 180^\circ$ From the figure, $\angle 2 + \angle 3 = \angle AOB$ and $\angle 6 + \angle 7 = \angle COD$. Thus, $\angle AOB + \angle COD = 180^\circ$.

Similarly, it can be shown that $\angle BOC + \angle DOA = 180^\circ$.

Final Answer: The opposite sides of the quadrilateral subtend supplementary angles at the center of the circle, as $\angle AOB + \angle COD = 180^\circ$ and $\angle BOC + \angle DOA = 180^\circ$.