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

Answer

To prove that the lengths of tangents drawn from an external point to a circle are equal, we will follow these steps:

### Step-by-Step Solution:

1. **Draw the Circle and Tangents**:
- Let \( O \) be the center of the circle with radius \( R \).
- Let \( P \) be an external point from which two tangents \( PA \) and \( PB \) are drawn to the circle, touching the circle at points \( A \) and \( B \) respectively.

2. **Identify the Right Angles**:
- According to the theorem of circles, the angle between the radius and the tangent at the point of contact is \( 90^\circ \).
- Therefore, \( \angle PAO = 90^\circ \) and \( \angle PBO = 90^\circ \).

3. **Consider the Triangles**:
- We will now consider triangles \( \triangle PAO \) and \( \triangle PBO \).

4. **Establish the Sides**:
- In triangle \( PAO \):
- \( OA = R \) (radius of the circle)
- In triangle \( PBO \):
- \( OB = R \) (radius of the circle)
- The side \( OP \) is common to both triangles, so \( OP = OP \).

5. **Apply the Criteria for Congruence**:
- We have:
- \( \angle PAO = \angle PBO = 90^\circ \)
- \( OA = OB = R \)
- \( OP = OP \) (common side)
- Therefore, by the RHS (Right angle-Hypotenuse-Side) criterion, we can conclude that:
\[ \triangle PAO \cong \triangle PBO \]

6. **Conclude using CPCTC**:
- By the property of congruent triangles (CPCTC - Corresponding Parts of Congruent Triangles are Congruent), we have:
\[ PA = PB \]
- Thus, the lengths of the tangents drawn from the external point \( P \) to the circle are equal.

### Final Statement:
Therefore, we have proved that the lengths of tangents drawn from an external point to a circle are equal, i.e., \( PA = PB \).

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