Question

in the adjoining figure two circles intersect at S and T. STP, BSC and BAP are straight lines. Prove that PATD is cyclic quadrilateral​

Answer 1

PATD is a cyclic quadrilateral

Step-by-step explanation:

In this figure as attachment -

∠1 is ∠PDT,  

∠2 is ∠TSC,  

∠3 is ∠PAT  

and ∠4 is ∠TSB

Also, ∠2 = ∠1

∠4= ∠3 (by exterior angle property of a cyclic quadrilateral)

We know that ∠4 + ∠2 is 180 by Linear pair Theorem

Therefore, ∠3+ ∠1 is 180

That means PATD is a cyclic quadrilateral

Therefore, P A and T are concyclic

Answer 2

please refer to the picture for your answer

Step-by-step explanation:

hopefully it helps you