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Consider a circle having the centre at O.
Let OA and OB are the 2 radii of the given circle.
Let
∠AOB=1300. Now, produce AO to X, such that AX is a diameter.
Also produce BO to Y, such that BY is a diameter.
Now, we will draw tangents at the end point A of diameter AX and end point B of diameter BY
Suppose the 2 tangents meet at point P outside the circle.
Now, AOBP becomes a quadrilateral.
We know that tangent drawn to a circle is always
⊥ to its radius at the point of contact.
Hence,
OA⊥AP and OB⊥BP. ⇒∠OAP=∠OBP=90∘ In quadrilateral AOBP,
∠APB+∠OBP+∠AOB+∠AOP=3600[Angle sum property]
⇒∠APB+900+1300+900=1800 ⇒∠APB=500 Now, in quadrilateral AOBP
∠OAP+∠OBP=900+900=1800 ∠APB+∠AOB=500+1300=1800 Now, we know that in a quadrilateral, if both the pairs of opposite angles are supplementary, then
Quadrilateral is a cyclic quadrilateral
⇒AOBP is a cyclic quadrilateral