### $AP$ and $BP$ are the two tangents at the extremities of chord $AB$ of a circle. Prove that $\angle MAP$ is equal to $\angle MBP$. A O P B M

Step by Step Explanation:
1. Given:
$AB$ is a chord of the circle with center $O$.
Tangents at the extremities of the chord $AB$ meet at an external point $P$.
Chord $AB$ intersects the line segment $OP$ at $M$.
2. Now, we have to find the measure of $\angle MAP.$

In $\triangle MAP$ and $\triangle MBP,$ we have \begin{aligned} & PA = PB && \text{[Tangents from an external point on a circle are equal in length]} \space \space\\ & MP = MP && \text{[Common]} \\ & \angle MPA = \angle MPB && \text { [Tangents from an external point are equally inclined to } \space \\ & && \text { the line segment joining the point to the center.] } \\ \implies & \triangle MAP \cong \triangle MBP && \text{ [by SAS Congruency Criterion] } \end{aligned}
3. We know that corresponding parts of congruent triangles are equal.
Thus, $\angle MAP = \angle MBP$.