Prove that the lengths of tangents drawn from an external point to a circle are equal.
Step-by-Step Solution
Key Concept: Draw tangents $PA, PB$ from external point $P$ to circle with center $O$. Prove $\Delta OAP \cong \Delta OBP$ by RHS.
Given: Tangents $PA$ and $PB$ from point $P$ to circle $C(O, r)$. To prove: $PA = PB$. [0.5 Mark]
Construction: Join $OA, OB, OP$. [0.5 Mark]
Proof: In right triangles $\Delta OAP$ and $\Delta OBP$:
1. $\angle OAP = \angle OBP = 90^\circ$ (Radius perpendicular to tangent)
2. $OA = OB = r$ (Radii of same circle)
3. $OP = OP$ (Common hypotenuse). [0.5 Mark]
By RHS congruence criterion, $\Delta OAP \cong \Delta OBP \Rightarrow PA = PB$. Proved! [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Given, To prove, and Construction: 0.5 Mark
Right angle setup $\angle OAP = \angle OBP = 90^\circ$: 0.5 Mark
RHS congruence step $\Delta OAP \cong \Delta OBP$: 0.5 Mark
Concluding $PA = PB$: 0.5 Mark
Correct Answer: