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ICSE - X
Arcs of a Circle
Lesson Demo
Theorem: The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the circle.
Given:
Arc AB.
Point C on the circle is outside AB.
To prove: ∠ AOB = 2 x ∠ ACB
Construction: Draw a line CO extended till point D.
Proof: In Δ OAC in each of these figures,
∠ AOD = ∠ OAC + ∠ OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)
OA = OC (Radii of same circle)
Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)
∠ AOD = ∠ OAC + ∠ OCA
⇒∠ AOD = 2 x ∠ OCA
Similarly, in Δ OBC,
∠ BOD = 2 x ∠ OCB
∠ AOD = 2 x ∠ OCA
and ∠ BOD = 2 x ∠ OCB
⇒∠ AOD + ∠ BOD = 2∠ OCA + 2∠ OCB
∠ AOD + ∠ BOD = 2 x (∠ OCA + ∠ OCB)
Or ∠ AOB = 2 x ∠ ACB
Hence, the theorem is proved.
Theorem: Angles subtended by an arc at all points within the same segment of the circle are equal.
Given:
An arc AB.
Points C and D are on the circle in the same segment.
To prove: ∠ ACB = ∠ ADB
Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the circle:
∠ AOB = 2 x ∠ ACB
Also, ∠ AOB = 2 x ∠ ADB
∴∠ ACB = ∠ ADB
Hence, the theorem is proved.
All angles formed in a semi circle are right angles
