Question

Solve any one of the following internal choices (a) or (b):
31(a) 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 proof is a very common 3-mark question in board exams.
Always draw a neat, labeled diagram showing the center $O$, tangent lines $PA$, $PB$, and chord $AB$ to make the steps easier to follow.

Answer 1

Step 1: Understanding the Question:
This is a standard proof-based question from the chapter "Circles".
We are given a circle with center $O$.
From an external point $P$, two tangents $PA$ and $PB$ are drawn to the circle, touching it at points $A$ and $B$ respectively.
We need to prove that the angle between the two tangents, $\angle APB$, is twice the angle between the chord $AB$ and the radius $OA$, which is $\angle OAB$.

Step 2: Key Formula or Approach:
1. Use the theorem: "The lengths of tangents drawn from an external point to a circle are equal." This means $PA = PB$, making $\Delta PAB$ an isosceles triangle.
2. Use the angle sum property of triangles to relate the angles of $\Delta PAB$.
3. Use the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact." This means $\angle OAP = 90^\circ$.

Step 3: Detailed Explanation:

• Let $\angle APB = \theta$.

• Since the lengths of tangents from an external point are equal:
\[ PA = PB \] Therefore, $\Delta PAB$ is an isosceles triangle.

• In an isosceles triangle, angles opposite to equal sides are equal:
\[ \angle PAB = \angle PBA \]

• Apply the angle sum property in $\Delta PAB$:
\[ \angle APB + \angle PAB + \angle PBA = 180^\circ \] \[ \theta + 2\angle PAB = 180^\circ \] \[ 2\angle PAB = 180^\circ - \theta \] \[ \angle PAB = 90^\circ - \frac{\theta}{2} \quad \text{--- (Equation 1)} \]

• Since the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$:
\[ \angle OAP = 90^\circ \]

• From the figure, we can write $\angle OAP$ as the sum of two adjacent angles:
\[ \angle OAB + \angle PAB = 90^\circ \]

• Substitute the value of $\angle PAB$ from Equation 1:
\[ \angle OAB + \left(90^\circ - \frac{\theta}{2}\right) = 90^\circ \] Subtract $90^\circ$ from both sides:
\[ \angle OAB - \frac{\theta}{2} = 0 \] \[ \angle OAB = \frac{\theta}{2} \] \[ \theta = 2 \angle OAB \]

• Substitute $\theta = \angle APB$ back into the relation:
\[ \angle APB = 2 \angle OAB \]

Step 4: Final Answer:
Hence Proved.