1. **Problem Statement:** Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. 2. **Setup and Notation:** Let the circle have center $O$ and radius $r$. Let $P$ be an external point from which two tangents $PA$ and $PB$ touch the circle at points $A$ and $B$ respectively. 3. **Key Properties:** - Tangents from an external point to a circle are equal in length: $PA = PB$. - Radius is perpendicular to tangent at the point of contact: $OA \perp PA$ and $OB \perp PB$. 4. **Goal:** Show that the angle between the tangents $\angle APB$ and the angle subtended by chord $AB$ at the center $\angle AOB$ satisfy: $$\angle APB + \angle AOB = 180^\circ$$ 5. **Proof:** - Since $OA$ and $OB$ are radii, triangle $OAB$ is isosceles with $OA = OB$. - Let $\angle OAB = \angle OBA = \theta$. - The angle at the center is $\angle AOB = 180^\circ - 2\theta$ (since sum of angles in triangle $OAB$ is $180^\circ$). 6. **Consider triangle $PAB$:** - Since $PA = PB$ (tangents from $P$), triangle $PAB$ is isosceles with $\angle PAB = \angle PBA = 90^\circ - \theta$ (because $OA \perp PA$ and $OB \perp PB$ imply $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$). - Therefore, the angle between the tangents at $P$ is: $$\angle APB = 180^\circ - 2(90^\circ - \theta) = 180^\circ - 180^\circ + 2\theta = 2\theta$$ 7. **Sum of angles:** $$\angle APB + \angle AOB = 2\theta + (180^\circ - 2\theta) = 180^\circ$$ 8. **Conclusion:** The angle between the two tangents from an external point is supplementary to the angle subtended by the chord joining the points of contact at the center of the circle.