### (a) Tangents from an external point are equal

**Given :** A circle with centre $O$; $P$ an external point; $PA$ and $PB$ are tangents touching the circle at $A$ and $B$.

**To prove :** $PA = PB$

**Construction :** Join $OA$, $OB$ and $OP$.

**Proof :**

**Step 1 —** $OA \perp PA$ and $OB \perp PB$ (radius $\perp$ tangent at point of contact)
$$\therefore\ \angle OAP = \angle OBP = 90^\circ$$

**Step 2 — In right triangles $\triangle OAP$ and $\triangle OBP$ :**
- $OA = OB$ (radii of the same circle)
- $OP = OP$ (common hypotenuse)
- $\angle OAP = \angle OBP = 90^\circ$

**Step 3 —** By **RHS congruence rule**,
$$\triangle OAP \cong \triangle OBP$$

**Step 4 —** By CPCT,
$$PA = PB$$

$$\textbf{Hence proved.}$$

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### (b) To prove $\angle PTQ = 2\angle OPQ$

**Step 1 — Let $\angle PTQ = \theta$.**

**Step 2 — Use equal tangents.**
$TP = TQ$ (tangents from external point $T$)
$\Rightarrow \triangle TPQ$ is isosceles $\Rightarrow \angle TPQ = \angle TQP$

**Step 3 — Apply the angle sum property in $\triangle TPQ$.**
$$\angle TPQ + \angle TQP + \angle PTQ = 180^\circ$$
$$2\angle TPQ = 180^\circ - \theta$$
$$\angle TPQ = 90^\circ - \frac{\theta}{2}$$

**Step 4 — Use the tangent–radius property at P.**
$$\angle OPT = 90^\circ$$

**Step 5 — Find $\angle OPQ$.**
$$\angle OPQ = \angle OPT - \angle TPQ = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}$$

**Step 6 — Conclude.**
$$\theta = 2\,\angle OPQ \ \Rightarrow\ \angle PTQ = 2\,\angle OPQ$$

$$\textbf{Hence proved.}$$