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:
The given information is geometrically represented as follows:
To prove:
\(AOBP\) is a square.
Proof:
Given that, \(\angle APB\) \(=\) .
By the theorem \(1\), we have:
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(OA\) \(\perp\) and \(OB\) \(\perp\) .
Thus, \(\angle OAP\) \(=\) \(\angle OBP\) \(=\) .
We know that:
The sum of all the angles in a quadrilateral is \(360^{\circ}\).
So, \(\angle APB\) \(+\) \(\angle OAP\) \(+\) \(\angle OBP\) \(+\) \(\angle AOB\) \(=\) \(360^{\circ}\)
\(\angle AOB\) \(=\)
By theorem \(2\), we have:
The lengths of tangents drawn from an exterior point to a circle are equal.
\(PA\) \(=\)
Also, \(OA\) and \(OB\) are equal (radius).
Here, all the four angles of the quadrilateral are equal.
Then, it is evident that all four sides are also equal.
Therefore, by the properties of the quadrilateral, we can conclude that it is a square.
Hence, proved.
Answer variants:
\(PA\)
\(PB\)
\(90^{\circ}\)