Question

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Hint:

This theorem also proves that \(OP\) bisects the angle between the two tangents (\(\angle APB\)) and the angle between the two radii (\(\angle AOB\)).

Answer 1

Concept Used:
A tangent to a circle is perpendicular to the radius at the point of contact. Also, triangles having a right angle, equal hypotenuse, and one equal side are congruent by RHS (Right angle–Hypotenuse–Side) criterion.

Proof:
1) Let a circle have centre $O$ and let $P$ be an external point.
From point $P$, draw two tangents $PA$ and $PB$ touching the circle at points $A$ and $B$ respectively.

2) Join $OA$, $OB$, and $OP$.

3) Since a radius is perpendicular to the tangent at the point of contact:
$OA \perp PA$ and $OB \perp PB$.
Therefore, $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.

4) In triangles $\triangle OAP$ and $\triangle OBP$:
- $OA = OB$ (Radii of the same circle).
- $OP = OP$ (Common side).
- $\angle OAP = \angle OBP = 90^\circ$.

5) Hence, $\triangle OAP \cong \triangle OBP$ by RHS congruence criterion.

6) By CPCT (Corresponding Parts of Congruent Triangles),
$PA = PB$.

Final Conclusion:
The lengths of tangents drawn from an external point to a circle are equal.