Question

Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that \(\angle\) APB = 2 \(\angle\) OAB.

Hint:

This is one of the most common circle theorems.
To remember the proof path, start by isolating the isosceles triangle formed by the external tangents.
Expressing its base angle in terms of \( \angle APB \) and subtracting it from the total \( 90^{\circ} \) angle of the tangent-radius perpendicularity will always lead directly to the solution.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Circles (Tangents and Angles properties).
We are given a circle with center \( O \) and two tangents \( PA \) and \( PB \) drawn from an external point \( P \).
We need to prove that the angle between the tangents, \( \angle APB \), is twice the angle \( \angle OAB \) formed by the chord \( AB \) and the radius \( OA \).

Step 2: Key Formula or Approach:
- Tangents drawn from an external point to a circle are equal: \( PA = PB \).
- In triangle \( \Delta PAB \), since two sides are equal, the base angles are equal: \( \angle PAB = \angle PBA \).
- The radius of a circle is perpendicular to the tangent at the point of contact: \( \angle OAP = 90^{\circ} \).

Step 3: Detailed Explanation:
1. Let \( \angle APB = \theta \).
2. Since \( PA \) and \( PB \) are tangents from the same external point \( P \), we have:
\[ PA = PB \]
Thus, \( \Delta PAB \) is an isosceles triangle.
3. In isosceles triangle \( \Delta PAB \), the angles opposite to the equal sides are equal:
\[ \angle PAB = \angle PBA \]
4. The sum of the angles in \( \Delta PAB \) is \( 180^{\circ} \):
\[ \angle APB + \angle PAB + \angle PBA = 180^{\circ} \]
\[ \theta + 2\angle PAB = 180^{\circ} \]
\[ 2\angle PAB = 180^{\circ} - \theta \]
\[ \angle PAB = 90^{\circ} - \frac{\theta}{2} \quad \text{(Equation 1)} \]
5. We know that the radius \( OA \) is perpendicular to the tangent \( PA \) at the point of contact \( A \):
\[ \angle OAP = 90^{\circ} \]
6. From the figure, we can express \( \angle OAP \) as:
\[ \angle OAP = \angle OAB + \angle PAB = 90^{\circ} \]
\[ \angle OAB = 90^{\circ} - \angle PAB \]
7. Substitute the value of \( \angle PAB \) from Equation 1:
\[ \angle OAB = 90^{\circ} - \left(90^{\circ} - \frac{\theta}{2}\right) \]
\[ \angle OAB = \frac{\theta}{2} \]
8. Multiply both sides by 2:
\[ 2\angle OAB = \theta \]
Substitute \( \theta = \angle APB \):
\[ \angle APB = 2\angle OAB \]
This completes the geometric proof.

Step 4: Final Answer:
Hence, it is proved that \(\angle APB = 2\angle OAB\).