Kuwait
Two tangents ^@ PA ^@ and ^@ PB ^@ are drawn to a circle with center ^@ O ^@ from an external point ^@ P ^@. Prove that ^@ \angle APB = 2 \angle OAB^@.
Answer:
- Given:
^@PA^@ and ^@PB^@ are the tangents to the circle with center ^@O^@. - Here, we have to find the value of ^@ \angle APB^@.
Let us consider ^@ \angle APB = x^\circ ^@.
We know that the tangents to a circle from an external point are equal.
So, ^@PA = PB ^@. - As ^@ \triangle APB ^@ is an isosceles triangle^@(PA = PB)^@, the base angles of the triangle will be equal. @^ \implies \angle PBA = \angle PAB @^
- We know that the sum of the angles of a triangle is ^@ 180^\circ ^@. Thus, @^ \begin{aligned} & \angle APB + \angle PAB + \angle PBA = 180^\circ \\ \implies & x^\circ + 2 \angle PAB = 180^\circ && \text{[As, } \angle PBA = \angle PAB] \\ \implies & \angle PAB = \dfrac { 1 }{ 2 } (180^\circ - x^\circ) = \bigg(90^\circ - \dfrac { 1 }{ 2 } x^\circ \bigg) \end{aligned} @^
- ^@ PA ^@ is a tangent and ^@ OA ^@ is the radius of the circle with center ^@ O ^@.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
@^ \begin{aligned} \implies & \angle OAP = 90^\circ \\ \implies & \angle OAB + \angle PAB = 90^\circ \\ \implies & \angle OAB = 90^\circ - (90^\circ - \dfrac { 1 }{ 2 } x^\circ) \\ \implies & \angle OAB = \dfrac { 1 }{ 2 } x^\circ = \dfrac { 1 }{ 2 } \angle APB \\ \implies & \angle APB = 2 \angle OAB \end{aligned} @^ - Hence, ^@ \angle APB = 2 \angle OAB.^@
