The pair of tangents \(AP\) and \(BP\) drawn from an external point \(P\) to a circle with centre \(O\) are perpendicular to each other. If so, then prove that the quadrilateral formed by the radii joining the ends of the tangents is a square.
Proof:
Given that, \(\angle APB\) \(=\) \(90^{\circ}\).
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(OA\) \(\perp\) \(PA\) and \(OB\) \(\perp\) \(PB\).
Thus, \(\angle OAP\) \(=\) \(\angle OBP\) \(=\) \(^{\circ}\).
The sum of all the angles in a quadrilateral is \(360^{\circ}\).
So, \(\angle APB\) \(+\) \(\angle OAP\) \(+\) \(\angle OBP\) \(+\) \(\angle AOB\) \(=\) \(^{\circ}\)
\(90^{\circ}\) \(+\) \(90^{\circ}\) \(+\) \(90^{\circ}\) \(+\) \(\angle AOB\) \(=\) \(360^{\circ}\)
\(270^{\circ}\) \(+\) \(\angle AOB\) \(=\) \(360^{\circ}\)
\(\angle AOB\) \(=\) \(360^{\circ}\) \(-\) \(270^{\circ}\)
\(\angle \) \(=\) \(90^{\circ}\)
The lengths of tangents drawn from an exterior point to a circle are equal.
\(PA\) \(=\) \(PB\)
Also, \(OA\) and are equal (radius).
Here, all four angles of the quadrilateral are equal.
Then, it is evident that all four sides are also equal.