arc theroem .Please explain it

1) The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
Given : Arc PQ of a circle C(O, r) and point r on the remaining part of the circle. 
Prove that: ∠ POQ = 2∠PRQ 
Construction: join RO and produce it to point M outside the circle.
 
Case 1: When arc PQ is a minor arc (figure (i))
Statements
Reasons
1) ∠POM = ∠OPM + ∠ORP 1) By exterior angle theorem (∠POM is exterior angle)
2) OP = OR 2) Radii of same circle
3) ∠OPR = ∠ORP 3) In a Δ two sides are equal then the angle opposite to them are also equal.
4) ∠POM = ∠ORP + ∠ORP 4) Substitution property. From (1)
5) ∠POM = 2∠ORP 5) Addition property.

Similarly, when ∠QOM is an exterior angle then ∠QOM = 2∠ORQ 
∴ from above, ∠POQ = 2∠PRQ

Case 2: When arc PQ is a semicircle.[figure(ii)]
Statements
Reasons
1) ∠POM = ∠OPR + ∠ORP 1) Exterior angle theorem.
2) ∠POM = ∠ORP + ∠ORP 2) As,OP = OR = radius. ∴∠ORQ = ∠ORP
3) ∠POM = 2∠ORP 3) Substitution and addition property.
4) ∠QOM = ∠ORQ + ∠OQR 4) Exterior angle theorem.
5) ∠QOM = ∠ORQ + ∠ORQ 5) As, OQ =OR = radius, ∴ ∠ORQ = ∠OQR
6) ∠QOM = 2∠ORQ 6) Substitution and addition property.

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