Arcs and Subtended Angles
In a circle, there are various angles that can be formed joining the endpoints of arcs and those angles are termed subtended angles. There are different categories of angles formed by these arcs, for example, angles in the same segment, angles in a semicircle, angles at the circumference, etc. Let's learn about it in detail in this lesson.
1.  How do you Find the Angle Subtended by an Arc in Circle? 
2.  Angle Subtended by an Arc at the Centre 
3.  Angles in the Same Segment of a Circle are Equal 
How do you Find the Angle Subtended by an Arc in Circle?
An arc of a circle is any part of the circumference. The angle subtended by an arc at any point is the angle formed between the two line segments joining that point to the endpoints of the arc. In the following figure, an arc of the circle shown subtends an angle α at a point on the circumference, and an angle β at the center O.
We are interested in the relation between the angle which a given arc subtends anywhere on the circumference and the angle which that arc subtends at the circle’s center. The relation between them is simple and extremely important.
Angle Subtended by an Arc at the Center
At the center of a circle, if there are two line segments originated from the endpoints of the arc of the circle intersect, that angle is called an angle subtended by the arc at the center. Let us understand a theorem and its proof based on it.
Theorem:
The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference.
The proof of this theorem is quite simple, and uses the exterior angle theorem – an exterior angle of a triangle is equal to the sum of the opposite interior angles. If the two opposite interior angles happen to be equal, then the exterior angle will be twice any of the opposite interior angles.
Proof:
Consider the following figure, in which an arc (or segment) AB subtends ∠AOB at the center O and ∠ACB at a point C on the circumference.
We have to prove that ∠AOB = 2 × ∠ACB. Draw the line through O and C, and let it intersect the circle again at D, as shown.
There are two triangles formed ΔOAC and ΔOBC. So, we make the following observations.
1. In ΔOAC, ∠OAC = ∠OCA, because OA = OC
2. In ΔOBC, ∠OBC = ∠OCB, because OB = OC.
Hence, using the exterior angle theorem, we get,
∠AOD= 2×∠ACO ⋯ (1)
∠DOB= 2×∠OCB ⋯ (2)
Add equations (1) and (2):
∠AOD+ ∠DOB= 2× (∠ACO+ ∠OCB)
⇒ ∠AOB= 2× ∠ACB
This completes the proof of the theorem. You can experiment with the simulation below to understand the relationship between the angles subtended by an arc at the center and at a point on the circle.
What if arc AB is such that it subtends a reflex angle at the center? Well, our proof doesn’t change, as the following figure shows.
You can now apply exactly the same steps as we did earlier and arrive at the same result: ∠AOB= 2× ∠ACB. Another configuration would be when O does not lie within ∠ACB. Will the result still hold, as in the following figure?
Yes, it will. Join C to O again and extend it to D on the circumference of the circle.
Use the Exterior Angle Theorem again and note that,
1. ∠BOD= 2× ∠BCD ⋯ (1)
2. ∠AOD= 2× ∠ACD ⋯ (2)
To avoid cluttering, these angles have not been highlighted in the figure, but make sure you observe them carefully. Subtract Equation (2) from (1)
∠BOD− ∠AOD= 2× (∠BCD− ∠ACD)
⇒ ∠AOB= 2× ∠ACB
This theorem leads to an interesting corollary, which is discussed next.
Corollary:
The angle in a semicircle is a right angle. Consider the figure below, where AB is the diameter of the circle. We need to prove that ∠ACB= 90^{∘}
Proof:
Since ∠AOB= 180^{º}, ∠ACB= 1/2× ∠AOB= 90^{º}
This theorem also leads to the following result.
Angles in the Same Segment of a Circle are Equal
In a circle, there are two segments major segments and minor segments. Angles formed by the endpoints of the chord in either major or minor segments are always equal. Let's understand this theorem and its proof in detail.
Theorem:
Angles in the same segment of a circle are equal. In other words, an arc in a circle will subtend equal angles anywhere on the circumference. Consider the following figure, which shows an arc AB subtending angles ACB and ADB at two arbitrary points C and D on the circumference. O is the center of the circle.
We need to prove that ∠ACB= ∠ADB.
Proof:
Using our previous theorem, we have that
∠ACB= 1/2× ∠AOB ⋯ (1)
∠ADB= 1/2× ∠AOB ⋯ (2)
From equations (1) and (2), we get ∠ACB= ∠ADB.
Important Notes:
 Congruent arcs of a circle subtend equal angles at the center.
 The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
 Angles in the same segment of a circle are equal.
 An angle in a semicircle is a right angle.
Think Tank:
 If the angles subtended by the chords of a circle at a center are equal, are the chords equal?
 The chords of congruent circles subtend equal angles at their centers. Are the chords equal?
Solved Examples

Example 1: In the following figure, if O is the center of the circle, find x^{º} in terms of y^{º} and z^{º}
Solution:
We label some angles as shown in the figure below.
Using the exterior angle theorem, we have,
∠y= ∠1+ ∠2 ⋯ (1)
∠1+ ∠x= ∠4 ⋯ (2)
Add Equations (1) and (2)
∠y+ ∠1+ ∠x= ∠1+ ∠2+ ∠4
⇒∠x+ ∠y= ∠2+ ∠4
Now, since z^{∘}= 2∠2= 2∠4, we have ∠2= ∠4= z^{∘}/2. So, x^{∘}+y^{∘}= z^{∘}, which implies x^{∘}= z^{∘}−y^{∘}.

Example 2: Two circles intersect at P and Q. Through P, diameters are drawn for both the circles. The other endpoints of these diameters are R and S. Show that R, Q, and S are collinear.
Solution:
Consider the following figure.
Since angles in semicircles are 90^{º}, we have, ∠RQP= ∠SQP= 90^{º}. Then, ∠RQS= ∠RQP+ ∠SQP, which can be written as 90^{∘}+90^{∘}, which is 180^{º}. Thus, RQS is a straight line, which means that R, Q and S collinear. Therefore, the points R, Q, and S are collinear.

Example 3: ΔABC is isosceles, with vertex A. Show that the circle through diameter AB bisects BC at point D.
Solution:
Observe the following figure. The circle cuts BC at D.
Since angles in a semicircle are of measure 90^{º}, ∠ADB= 90^{∘}. This means AD is the altitude for the isosceles ΔABC. This implies, AD also bisects BC. Therefore, D is the midpoint of BC.
FAQs on Arcs and Subtended Angles
What is the Meaning of Subtended?
In Geometry, subtended means to extend under. A subtended angle is formed by the intersection of more than 1 line, or the rays joining the endpoints of a line or arc.
What is a Subtended Central Angle?
A subtended central angle is the angle formed at the center of the circle and the legs of the angle are the radii of the circle intersecting the circle at two different points.
How do you Find the Angle Subtended by an Arc at the Centre?
To find the value of angle subtended by an arc at the center we have to multiply the angle formed through the same endpoints of the arc on the circumference by two. For example, if the angle subtended at any point on the circumference is 60^{º}, that means the angle subtended by the same arc at the center is 120^{º}.
What is the Central Angle of a Circle?
The central angle of a circle is that angle whose vertex is the same as the center point of the circle. The other two arms are the radius which meets the endpoints of any chord or arc of the circle.
What is Arc in Circle?
In a circle, the arc can be defined as any portion of the boundary. It is a curve that forms the circumference of a circle. In other words, we can say that circumference is a complete or closed arc of the circle.
What is the Measure of Angle in a Semicircle?
The angle in a semicircle is always 90^{º}. Arc drawn from the endpoints of the diameter of a circle always form a 90^{º }angle on its circumference.