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))
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)]
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))
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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)]
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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. |