Question

In given figure, RS is diameter and PQ chord of a circle with centre O. Prove that (a) ∠RPO = ∠OQR (b) ∠POQ = 2∠PRO

Answer 1

Hence it is proved that,

(a) ∠RPO = ∠OQR (b) ∠POQ = 2∠PRO

Let RS intersect PQ at M.

Now,

In △PRM and △QRM

⇒ RM = RM                                (Common Side)

⇒ ∠RMP = ​∠RMQ = 90       (A line from center of the circle is a perpendicular bisector of the chord of the same circle)  

⇒PM = MQ    (A line from center of the circle is a perpendicular bisector of the chord of the same circle)  

Therefore, ​△PRM ≅ △QRM  by SAS test

Hence, PR = QR         .......................... (1)

a) In △RPO and △RQO

⇒PO = QO                      (radius of the circle)

⇒RO = RO                     (Common side)

⇒PR = QR                      (From (1) )

Therefore,  △RPO ≅ △RQO  by SSS test

b) ∠POQ = 2∠PRO

As we know that the angle subtended at the center of the circle is twice the angle subtended at any other point on the circle,

therefore, we have,

∠ POQ = 2 ∠ PRO

where, ∠ POQ = angle subtended at the center of the circle

∠ PRO = angle subtended at a point R on the circle.

Answer 2

Answer:

see the attached and u will get the answer