Question

can anybody pls xplain theorum 10.10 of chapter circles of class 9

Answer 1

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'QThis 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.

Answer 2

I think the ans of theorem 10.10 is the reverse of theorem 10.9