PQ is a diameter. The tangent drawn at the point R, intersects the two tangents drawn at the points P and Q at the points A and B respectively. Prove that `angle AOB` is a right angle.

Explanation

Let us join O, R. Also, let OA intersects the are PR at E and OB intersects the are PQ at F.
`because AP = AR, therefore` PE = are RE
The central angles subtended by then are equal, i.e., `anglePOE` f
`= ange ROE or, angle AOP = angle AOR`
Similarly. `angle BOR=angle BOQ.`
Then, `angle AOP+ angle AOR + angle BOR+ angle BOQ =180^(@)[because` straight angle]
or, `2 angle AOR +2 angle BOR =180^(@) or, 2 (angle AOR+ angle BOR)=180^(@)`
or, `2 angle AOB=180^(@)or, 2 (angle AOR+angleBOR=180^(@)`
or,` 2 angle AOB=180^(@)or, angle AOB=(180^(@))/(2)=90^(@)`
Hence `angleAOB` is a right angle. (Proved)