Problem Statement
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Verified Solution & Marking Scheme
Given, To Prove, and Construction
**Given:** A circle with center $O$, 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 $OR$.
Establish right angles at points of tangency
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, $OQ \perp PQ \implies \angle OQP = 90^\circ$.
Similarly, $OR \perp PR \implies \angle ORP = 90^\circ$.
Prove congruence of right triangles △OQP and △ORP
In right $\triangle OQP$ and right $\triangle ORP$:
1. $\angle OQP = \angle ORP = 90^\circ$ (Right angles)
2. $OP = OP$ (Common hypotenuse)
3. $OQ = OR$ (Radii of the same circle)
By RHS congruence criterion: $\triangle OQP \cong \triangle ORP$
Conclude via CPCT
$PQ = PR \quad (\text{Corresponding Parts of Congruent Triangles})$
Hence proved.