PQRS is a cyclic quadrilateral. Angle SPT is an exterior angle. SP is a bisector of angle RPT and PQRS is a cyclic quadrilaterla.

Prove SQ=SR

Given : PQRS is a cyclic quadrilateral 

SP is bisector of ∠RPT

To prove : SQ = SR

Construction : Join SQ

So, it is sufficient to prove ∠3 = ∠4

Proof : SP is bisector of ∠RPT

⇒ ∠1 = ∠2       ...(1)

We know that angle in a same segment are equal.

∴ ∠1 = ∠4 and ∠5 = ∠6  ...(2)

Now ∠1 + ∠2 + ∠5 = 180°  ...(3)

Also, in ΔRQS, ∠3 + ∠4 + ∠6 = 180°  ...(4)  (by angle sum property of Δ)

equating (1) and (2)

∠1 + ∠2 + ∠5 = ∠3 + ∠4 + ∠6

⇒ ∠2 = ∠3  (using (2))

But from (1), ∠1 = ∠2

⇒∠1 = ∠2 = ∠3 = ∠4  (using (3))

⇒ ∠3 = ∠4

∴ SQ = SR  (Sides opposite to equal angles are equal)