Loading....
Coupon Accepted Successfully!

 

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 PRQ = 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
PRQ = PSQ (given)  (i)
And
PRQ = PS'Q  [Angles in the same segment]  (ii)
∴ ∠PSQ = 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.

 

Example

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

Solution

In the regular pentagon ABCDE, join BD and CE.
In
ΔBCD and ΔCDE
CD = CD
BC = DE

BCD = CDE
∴ΔBCD ≅ ΔCDE
\∠1 = 2
But they are subtended on the same side of CD and by CD.

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


 





Test Your Skills Now!
Take a Quiz now
Reviewer Name