PQ is a diameter of the circle with centre O. The tangent drawn at any point R on the circle intersects the tangents drawn at P and Q at two points A and B respectively. Prove that `angle AOB =1` right angle.

Explanation

Let PQ is a dimetr of the circle with centre O. EF and CD are two of its tangents drawn at P and Q respectively. R lies on the circle and the tangent AB drawn at R intersects the previous tangents at A and B respectively. Let us join O, A, O, B.
To prove : `angle AOB =1` right angle.
Proof : In `DeltaAOPand DeltaAOR,`
`angle APO =angle A RO [because ` ech is right angle]
`OR =OP [because` radii of same circle ]
and hypotenuse OA is common to both.
`therefore DeltaAOP -= DeltaAOR` [by the RHS condition of congruency]
`therefore angle OAP=angle OAR......(1)[because` similar angles of two congruent triangles]
Similarly, it can be proved that `angle OBR=angle OBQ......(2)`
Then, `angle AOB=angle AOR+angleBOR`
`=90^(@) -angle OAR+90^(@) -angle OBR`
`=90^(@) - angle OAP +90^(@)-angleOBQ`
`=angle AOP +angle BOQ`
`=180^(@) -anlgeAOB [becauseangle AOB+ angle AOP+angle BOQ=180^(@)]`
or `2 angle AOB =180^(@)or , angle AOB=90^(@)`
Hence `angle AOB=1` right angle. (Proved)