Russia
^@ PA ^@ and ^@ PB ^@ are tangents to a circle with center ^@ O ^@ from an external point ^@ P ^@, touching the circle at ^@ A ^@ and ^@ B ^@ respectively. Show that the quadrilateral ^@ AOBP ^@ is cyclic.
Answer:
- Given:
^@ PA ^@ and ^@ PB ^@ are the tangents to the circle with center ^@ O ^@ from an external point ^@ P ^@.
Here, we have to check if quadrilateral ^@ AOBP ^@ is cyclic or not.
We know that in a cyclic quadrilateral the sum of opposite angles is ^@ 180^\circ^@. - Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Thus, @^ \begin{aligned} & PA \perp OA \implies \angle OAP = 90^\circ \\ & \text{ and } \\ & PB \perp OB \implies \angle OBP = 90^\circ \end{aligned} @^ So, @^ \angle OAP + \angle OBP = 90^\circ + 90^\circ = 180^\circ \ldots \text{(i)}@^ - Now, the sum of all the angles of a quadrilateral is ^@ 360^\circ^@. Thus, @^ \begin{aligned} & \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \\ & \angle AOB + \angle APB + ( \angle OAP + \angle OBP )= 360^\circ \\ \implies & \angle AOB + \angle APB = 180^\circ && \text{ [Using } eq \text{(i)] } \end{aligned} @^
- Both pairs of opposite angles have the sum ^@ 180^\circ ^@. Thus, we can say that quadrilateral ^@ AOBP ^@ is ^@ cyclic.^@
