Lesson 2, Topic 5
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# Angles in the Same Segment of a Circle

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Theorem: Angles in the same segment are equal.

Given: Consider the circle as shown above, in which chord 𝑃𝑄 subtends angles 𝑃𝐴𝑄 and 𝑃𝐶𝑄 at any two points 𝐴 and 𝐶 on the circumference of a circle.

To prove:𝑃𝐴𝑄 = ∠𝑃𝐶𝑄

### Proof:

In the last lesson, we learnt that the angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

Thus, ∠𝑃𝑂𝑄 = 2∠𝑃𝐴𝑄…….(𝑖)

Similarly, ∠𝑃𝑂𝑄 = 2∠𝑃𝐶𝑄…….(𝑖𝑖)

Equate (𝑖) and (𝑖𝑖)

2∠𝑃𝐴𝑄 = 2∠𝑃𝐶𝑄

∴ ∠𝑃𝐴𝑄 = ∠𝑃𝐶𝑄

Therefore angles in the same segments of a circle are equal.

These angles can be more than two. Therefore, we can say that all angles in the major segment are equal.

For the major segment in the circle below all the angles are equal.

∴ ∠a = ∠b = ∠c = ∠d

Similarly, all angles in a minor segment are all equal.

For the minor segment in the circle below all the angles are equal.

∴ ∠x = ∠y = ∠z

### Example

In the figure below ∠𝐵𝐴𝐷 = 36° and ∠𝐶𝐵𝐴 = 37°, find ∠𝐵𝐶𝐷 and ∠𝐶𝐷𝐴.

Solution:

Let’s add the given angles to the diagram.

If we draw lines to join BD and AC, we get two chords that separate the circle into segments.

If we consider the major segment of chord BD

∠𝐵𝐶𝐷 = ∠𝐵𝐴𝐷 = 36° (angles in the same segment)

If we consider the major segment of chord AC

∠𝐶𝐷𝐴 = ∠𝐶𝐵𝐴 = 37° (angles in the same segment)

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