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

Step 1: Understanding the Concept:
We use the property of congruent triangles. A tangent is always perpendicular to the radius at the point of contact.
Step 2: Key Formula or Approach:
RHS (Right angle-Hypotenuse-Side) congruence criterion.
Step 3: Detailed Explanation:
1. Given: A circle with center \(O\) and a point \(P\) outside the circle. \(PA\) and \(PB\) are two tangents.
2. To Prove: \(PA = PB\).
3. Construction: Join \(OA, OB, \text{ and } OP\).
4. Proof: In \(\triangle OAP\) and \(\triangle OBP\):
- \(\angle OAP = \angle OBP = 90^\circ\) (Radius is \(\perp\) to tangent).
- \(OP = OP\) (Common hypotenuse).
- \(OA = OB\) (Radii of the same circle).
5. By RHS congruence criterion, \(\triangle OAP \cong \triangle OBP\).
6. Therefore, \(PA = PB\) by CPCT (Corresponding Parts of Congruent Triangles).
Step 4: Final Answer:
The lengths of tangents drawn from an external point to a circle are equal.