If the sum of a pair …

Given the sum of pair of opposite angles of a quadrilateral is 180° we have to prove that the quadrilateral is cyclic.

Let us assume that the quadrilateral ABCD is not cyclic i.e Let the point D does not lie on the circle which makes the quadrilateral non-cyclic. Now, let us do a construction such that join CD' where D' is the point of intersection of side AD with the circle.

Now, ABCD' is cyclic

⇒  ∠3 +  ∠4 = 180°

Now, it is given that  the sum of pair opposite angles of a quadrilateral ABCD is 180°

Therefore, ∠2 +  ∠4 = 180°

From above two equations we get

∠3 +  ∠4 =  ∠2 +  ∠4

⇒  ∠3 = ∠2

Now, in triangle CDD', by external angle property

∠3  = ∠1 + ∠2

⇒ ∠1 = 0 , hence the side CD' and CD coincides

⇒ Point D lies on circle

Hence, our supposition is wrong quadrilateral ABCD is cyclic.