Question

In the given Fig. O is the center of the circle, chord PQ is parallel and equal to chord RS and QR is the diameter. Prove that arc PR = arc QS.

Answer 1

Join PS 

Now, ∠POR = 2 ∠PQR (∠ at the center in twice the ∠ at the remaining circumference) 

∠SOQ = 2 ∠SRQ (∠ at the center is twice the ∠ at the remaining circumference) 

But ∠PQR = ∠SRQ (Alternate ∠s, PQ || RS) 

∴ ∠POR = ∠SOQ

⇒ Arc PR = Arc QS (In the same circle, arcs subtending equal angles at the center are equal.) Hence, proved. 

[Note. It is clear otherwise also.]

 ∵ PQR = SRQ 

∴arc PR = arc QS.