Canada
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^@.
Answer:
^@ 90^\circ ^@
- Given:
^@ PA ^@ and ^@ PB ^@ are the tangents to the circle from an external point ^@ P ^@.
To find:
The measure of ^@\angle AMP ^@. - 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]} ^@
- 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} @^ - Therefore, the measure of ^@ \angle AMP ^@ is ^@ 90^\circ ^@.
