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A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent.

Corresponding arcs of two equal chords of a circle are congruent.

**Congruent arcs of a circle subtend equal angles at the centre.**

Given: Two congruent arcs AB and CD.

To prove: âˆ AOB = âˆ COD

Construction: Draw chords AB and CD.

Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre.

In the given figure,

AB = CD (Chords corresponding to congruent arcs of a circle are equal)

âˆ AOB = âˆ COD (Equal chords subtend equal angles at the centre)

Hence, the theorem is proved.

**The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
**

Given: Arc AB. Point C on the circle is outside AB.

To prove: âˆ AOB = 2 Ã— âˆ ACB

Construction: Draw a line CO extended till point D.

Proof: In Î”OAC in each of these figures,

âˆ AOD = âˆ OAC + âˆ OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)

OA = OC (Radii of same circle)

Thus, âˆ OAC = âˆ OCA (Angles opposite equal sides of a triangle are equal)

âˆ AOD = âˆ OAC + âˆ OCA

â‡’âˆ AOD = 2 Ã— âˆ OCA

Similarly, in Î”OBC, âˆ BOD = 2 Ã— âˆ OCB

âˆ AOD = 2 Ã— âˆ OCA and âˆ BOD = 2 Ã— âˆ OCB

â‡’ âˆ AOD + âˆ BOD = 2 âˆ OCA + 2 âˆ OCB

âˆ AOD + âˆ BOD = 2 Ã— (âˆ OCA + âˆ OCB)

or âˆ AOB = 2 Ã— âˆ ACB

Hence, the theorem is proved.

Given: An arc AB. Points C and D are on the circle in the same segment.

To prove: âˆ ACB = âˆ ADB

Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle:

âˆ AOB = 2 Ã— âˆ ACB

Also, âˆ AOB = 2 Ã— âˆ ADB

âˆ´ âˆ ACB = âˆ ADB

Hence, the theorem is proved.

Given: A circle with centre O, and Q, P and R are three points on the circumference of the circle.

Construction: Join the points Q, P and R to the points A and B.

To prove: âˆ AQB = âˆ APB = âˆ ARB

Proof: âˆ AQB = $\frac{\text{1}}{\text{2}}$ âˆ AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

âˆ APB = $\frac{\text{1}}{\text{2}}$ âˆ AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

Similarly, âˆ ARB = $\frac{\text{1}}{\text{2}}$ âˆ AOB.

âˆ´ âˆ AQB = âˆ APB = âˆ ARB

Hence, the theorem is proved.

A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent.

Corresponding arcs of two equal chords of a circle are congruent.

**Congruent arcs of a circle subtend equal angles at the centre.**

Given: Two congruent arcs AB and CD.

To prove: âˆ AOB = âˆ COD

Construction: Draw chords AB and CD.

Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre.

In the given figure,

AB = CD (Chords corresponding to congruent arcs of a circle are equal)

âˆ AOB = âˆ COD (Equal chords subtend equal angles at the centre)

Hence, the theorem is proved.

**The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
**

Given: Arc AB. Point C on the circle is outside AB.

To prove: âˆ AOB = 2 Ã— âˆ ACB

Construction: Draw a line CO extended till point D.

Proof: In Î”OAC in each of these figures,

âˆ AOD = âˆ OAC + âˆ OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)

OA = OC (Radii of same circle)

Thus, âˆ OAC = âˆ OCA (Angles opposite equal sides of a triangle are equal)

âˆ AOD = âˆ OAC + âˆ OCA

â‡’âˆ AOD = 2 Ã— âˆ OCA

Similarly, in Î”OBC, âˆ BOD = 2 Ã— âˆ OCB

âˆ AOD = 2 Ã— âˆ OCA and âˆ BOD = 2 Ã— âˆ OCB

â‡’ âˆ AOD + âˆ BOD = 2 âˆ OCA + 2 âˆ OCB

âˆ AOD + âˆ BOD = 2 Ã— (âˆ OCA + âˆ OCB)

or âˆ AOB = 2 Ã— âˆ ACB

Hence, the theorem is proved.

Given: An arc AB. Points C and D are on the circle in the same segment.

To prove: âˆ ACB = âˆ ADB

Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle:

âˆ AOB = 2 Ã— âˆ ACB

Also, âˆ AOB = 2 Ã— âˆ ADB

âˆ´ âˆ ACB = âˆ ADB

Hence, the theorem is proved.

Given: A circle with centre O, and Q, P and R are three points on the circumference of the circle.

Construction: Join the points Q, P and R to the points A and B.

To prove: âˆ AQB = âˆ APB = âˆ ARB

Proof: âˆ AQB = $\frac{\text{1}}{\text{2}}$ âˆ AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

âˆ APB = $\frac{\text{1}}{\text{2}}$ âˆ AOB [Angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.]

Similarly, âˆ ARB = $\frac{\text{1}}{\text{2}}$ âˆ AOB.

âˆ´ âˆ AQB = âˆ APB = âˆ ARB

Hence, the theorem is proved.