Two tangents \(PA\) and \(PB\) are drawn from an external point \(P\) to the circle with centre \(O\). If the tangents are perpendicular to each other, prove that the quadrilateral \(OAPB\) is a square.
Proof:
Given that, \(\angle APB\) \(=\) \(90^{\circ}\).
By the theorem \(1\), we have:
\(OA\) \(\perp\) \(PA\) and \(OB\) \(\perp\) \(PB\).
Thus, \(\angle OAP\) \(=\) \(\angle OBP\) \(=\) \(^\circ\)
We know that:
The sum of all the angles in a quadrilateral is \(^\circ\)
So, \(\angle APB\) \(+\) \(\angle OAP\) \(+\) \(\angle OBP\) \(+\) \(\angle AOB\) \(=\) \(^\circ\)
By simplifying we get, \(\angle AOB\) \(=\) \(^\circ\)
By theorem \(2\), we have:
\(PA\) \(=\)
Also, \(OA\) and are equal (radius).
Here, all the four angles of the quadrilateral are .
Then, it is evident that all four sides are also .
Therefore, by the properties of the quadrilateral, we can conclude that it is a square.
Hence, proved.