From a point $P$ outside a circle with center $O$, tangents $PA$ and $PB$ are drawn to the circle. The line segment formed by joining the points $A$ and $B$ intersect the line segment $OP$ at $M$. Find the measure of $\angle AMP$. A O P B M

$90^\circ$

Step by Step Explanation:
1. Given:
$PA$ and $PB$ are the tangents to the circle from an external point $P$.

To find:
The measure of $\angle AMP$.
2. In $\triangle MAP$ and $\triangle MBP$, we have \begin{aligned} & PA = PB && \text{[ Tangents to a circle from an external point are equal.]} \\ & MP = MP && \text{[Common side]} \\ & \angle OPA = \angle OPB && \text{[Tangents from an external point are equally inclined to } \space\space\\ & && \text { the line segment joining the center to that point.]} \\ \implies & \angle MPA = \angle MPB && \text{[As, } \angle OPA = \angle MPA \text{ and } \angle OPB = \angle MPB ] \end{aligned} Thus, $\triangle MAP \cong \triangle MBP \space \space \space \text{[By SAS-congruence]}$
3. As the corresponding parts of congruent triangles are equal, $MA = MB$ and $\angle AMP = \angle BMP.$

Also, \begin{aligned} \ & \angle AMP + \angle BMP = 180^\circ && \text{[Angles on a straight line.]} \\ \implies & \angle AMP + \angle AMP = 180^\circ && \text{[As, } \angle AMP = \angle BMP] \\ \implies & 2 \angle AMP = 180^\circ \\ \implies & \angle AMP = \dfrac { 180^\circ } { 2 } = 90^\circ \\ \implies & \angle AMP = \angle BMP = 90^\circ \end{aligned}
4. Therefore, the measure of $\angle AMP$ is $90^\circ$.