Q10:

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.

Given: A circle with center $O$. An external point $P$ from which two tangents $PA$ and $PB$ are drawn to the circle, touching the circle at points $A$ and $B$ respectively. The line segment $AB$ joins the points of contact.

To Prove: $\angle APB + \angle AOB = 180^\circ$ (i.e., they are supplementary).

Step 1: Identify the properties of tangents to a circle.

According to the theorem: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OA \perp PA$ and $OB \perp PB$.

This implies that $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$.

Step 2: Consider the quadrilateral $OAPB$.

In the quadrilateral $OAPB$, the sum of the interior angles is given by the angle sum property of a quadrilateral.

$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^\circ$ [Sum of interior angles of a quadrilateral is $360^\circ$].

Step 3: Substitute the known values into the equation.

Substitute $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$ into the equation:

$90^\circ + \angle APB + 90^\circ + \angle AOB = 360^\circ$

Step 4: Simplify the algebraic expression.

Combine the constant terms:

$180^\circ + \angle APB + \angle AOB = 360^\circ$

Subtract $180^\circ$ from both sides of the equation:

$\angle APB + \angle AOB = 360^\circ - 180^\circ$

$\angle APB + \angle AOB = 180^\circ$

Conclusion:

Since the sum of $\angle APB$ and $\angle AOB$ is $180^\circ$, the angle between the two tangents and the angle subtended by the line segment joining the points of contact at the center are supplementary.

Final Answer: Hence, it is proved that $\angle APB + \angle AOB = 180^\circ$.