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: Subdivide 8 angles around centre $O$. $\angle 1 = \angle 2, \angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$. Total sum $= 360^\circ \Rightarrow 2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ \Rightarrow \angle AOB + \angle COD = 180^\circ$.
Join centre $O$ to all 4 vertices and 4 points of contact. 8 congruent right triangles formed. [1.0 Mark]
Angles around centre: $\angle 1 = \angle 2, \angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$. [1.0 Mark]
$2(\angle 2 + \angle 3 + \angle 6 + \angle 7) = 360^\circ \Rightarrow \angle AOB + \angle COD = 180^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Subdividing 8 central angles: 1.0 Mark
Equating congruent central angle pairs: 1.0 Mark
Concluding $\angle AOB + \angle COD = 180^\circ$: 1.0 Mark
Correct Answer: