Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Step-by-Step Solution
Key Concept: Let $AB$ be diameter. Tangent $L_1$ at $A \Rightarrow \angle OAL_1 = 90^\circ$. Tangent $L_2$ at $B \Rightarrow \angle OBL_2 = 90^\circ$. Alternate interior angles equal $\Rightarrow L_1 \parallel L_2$.
$\angle OAL_1 = 90^\circ$ and $\angle OBL_2 = 90^\circ$ (Radius $\perp$ Tangent). [1.0 Mark]
$\angle OAL_1 = \angle OBL_2$ (Alternate interior angles) $\Rightarrow L_1 \parallel L_2$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Stating right angles at points of contact: 1.0 Mark
Equating alternate interior angles to show parallel: 1.0 Mark
Correct Answer: