Kuwait
^@ 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 ^@.
Answer:
- 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^@. - 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} @^ - We know that corresponding parts of congruent triangles are equal.
Thus, ^@ \angle MAP = \angle MBP ^@.
