Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Step-by-Step Solution
Key Concept: Join center $O$ to all 4 vertices and 4 points of contact, showing 4 pairs of equal congruent triangles.
Let $ABCD$ circumscribe circle $C(O,r)$ touching at $P,Q,R,S$. Join $OP, OQ, OR, OS$ and $OA, OB, OC, OD$. [0.5 Mark]
$\Delta OAP \cong \Delta OAS \Rightarrow \angle 1 = \angle 8$.
$\Delta OBP \cong \Delta OBQ \Rightarrow \angle 2 = \angle 3$.
$\Delta OCQ \cong \Delta OCR \Rightarrow \angle 4 = \angle 5$.
$\Delta ODR \cong \Delta ODS \Rightarrow \angle 6 = \angle 7$. [1.5 Marks]
Sum around center $= 360^\circ \Rightarrow 2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^\circ \Rightarrow \angle AOB + \angle COD = 180^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Construction and vertex/contact line connections: 0.5 Mark
Establishing 4 angle equality pairs via congruent triangles: 1.5 Marks
Adding angles around center to prove $\angle AOB + \angle COD = 180^\circ$: 1.0 Mark
Correct Answer: