In the given figure, $XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^\circ$.
Step-by-Step Solution
Key Concept: $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA$. $\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB$. Since $POQ$ is straight line ($180^\circ$), $2\angle COA + 2\angle COB = 180^\circ \Rightarrow \angle AOB = 90^\circ$.
Join $OC$. $\Delta OPA \cong \Delta OCA \Rightarrow \angle POA = \angle COA$. (1) [1.0 Mark]
$\Delta OQB \cong \Delta OCB \Rightarrow \angle QOB = \angle COB$. (2) [1.0 Mark]
$POQ$ is a diameter line $\Rightarrow \angle POA + \angle COA + \angle COB + \angle QOB = 180^\circ \Rightarrow 2\angle COA + 2\angle COB = 180^\circ \Rightarrow \angle AOB = 90^\circ$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving congruence of upper triangles: 1.0 Mark
Proving congruence of lower triangles: 1.0 Mark
Summing angles to prove $\angle AOB = 90^\circ$: 1.0 Mark
Correct Answer: