Problem Statement
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Verified Solution & Marking Scheme
State Given, To Prove, and Construction
- **Given:** A circle with centre $O$, and a point $P$ lying outside the circle. $PQ$ and $PR$ are two tangents drawn from $P$ touching the circle at $Q$ and $R$ respectively.
- **To Prove:** $PQ = PR$.
- **Construction:** Join $OP$, $OQ$, and $OR$.
Identify Right Angles at Points of Contact
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore:
$\angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ$
Apply RHS Congruence Criterion
In right triangles $\triangle OQP$ and $\triangle ORP$:
1. $\angle OQP = \angle ORP = 90^\circ$ (Right angles)
2. $OP = OP$ (Common hypotenuse)
3. $OQ = OR$ (Radii of the same circle)
Therefore, by RHS congruence criterion:
$\triangle OQP \cong \triangle ORP$
Conclude via CPCTC
Since corresponding parts of congruent triangles are equal (CPCTC):
$PQ = PR$
Hence proved.