# Converse of Theorem 6

**Converse : ** If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, the four points are concyclic.

**Example :** lie on the same circle.

**Given**** :** PQ is a line segment and R, S are two points lying on the same side of the line containing PQ, such that âˆ PRQ = âˆ PSQ.

**To Prove :** P, Q, R and S are concyclic.

**Construction :** Draw a circle through the three non-collinear points P, Q, R.

**Proof :** If we assume that point S does not lie on the circle, then the circle will intersect the line PS at a point 'say S'.

Now âˆ PRQ = âˆ PSQ (given) (i)

And âˆ PRQ = âˆ PS'Q [Angles in the same segment] (ii)

âˆ´ âˆ PSQ = âˆ PS'Q

This is possible only when S lies on S', and when S coincides with S'. Thus our assumption that S does not lie on the circle is false. Hence P, Q, R and S are concyclic.

Prove that any four vertices of a regular pentagon are concyclic.

In the regular pentagon ABCDE, join BD and CE.

In Î”BCD and Î”CDE

CD = CD

BC = DE

âˆ BCD = âˆ CDE

âˆ´Î”BCD â‰… Î”CDE

\âˆ 1 = âˆ 2

But they are subtended on the same side of CD and by CD.

âˆ´ Points B, C, D and E are concyclic. [angles on the same chord of a circle are equal in the same segment]