In given figure, RS is diameter and
PQ chord of a circle with centre O.
Prove that (a)
∠
RPO =
∠
OQR (b)
∠
POQ = 2
∠
PRO
Answers
Answer:
Hence it is proved that,
(a) ∠RPO = ∠OQR (b) ∠POQ = 2∠PRO
∠ PRO = angle subtended at a point R on the circle.
Step-by-step explanation:
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.