Let `P Q` and `R S` be tangent at the extremities of the diameter `P R` of a circle of radius `r`. If `P`S and `R Q` intersect at a point `X` on the circumference of the circle, then prove that `2r=sqrt(P Q xx R S)` .

Explanation

To prove that \(2r = \sqrt{PQ \cdot RS}\), we will follow these steps:

1. **Understanding the Setup**: We have a circle with diameter \(PR\) and radius \(r\). The points \(P\) and \(R\) are at the ends of the diameter. The tangents \(PQ\) and \(RS\) touch the circle at points \(P\) and \(R\) respectively. The lines \(PS\) and \(RQ\) intersect at point \(X\) on the circumference of the circle.

2. **Identifying Angles**: Since \(PR\) is the diameter of the circle, the angle \(PXQ\) is a right angle (90 degrees) due to the property of a semicircle. Let’s denote the angle \(PXQ\) as \(90^\circ\).

3. **Using Right Triangle Properties**: In triangle \(PQR\), we can apply the tangent function:
\[
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