If the two equal chords of a circle intersect :
(i) inside
(ii) on
(iii) outside
the circle, then show that the line segment joining the point of intersection to the centre of the circle will bisect the angle between the chords.

Explanation

To solve the problem of showing that the line segment joining the point of intersection of two equal chords to the center of the circle bisects the angle between the chords, we will consider three cases: (i) when the chords intersect inside the circle, (ii) when they intersect on the circle, and (iii) when they intersect outside the circle.

### Step-by-Step Solution:

#### Case (i): Chords intersect inside the circle

1. **Draw the Circle and Chords**:
- Draw a circle and label its center as \( O \).
- Draw two equal chords \( AB \) and \( CD \) that intersect at point \( X \) inside the circle.

2. **Draw Perpendiculars from the Center**:
- From point \( O \), draw perpendiculars to the chords \( AB \) and \( CD \) at points \( M \) and \( N \) respectively.

3. **Identify Triangles**:
- Consider triangles \( \triangle OXM \) and \( \triangle OXN \).

4. **Establish Congruence**:
- In triangles \( OXM \) and \( OXN \):
- \( OX = OX \) (common side),
- \( \angle OMX = \angle ONX = 90^\circ \) (perpendiculars),
- \( OM = ON \) (equal distance from center to equal chords).
- By RHS (Right angle-Hypotenuse-Side) congruence, \( \triangle OXM \cong \triangle OXN \).

5. **Use CPCT**:
- By CPCT (Corresponding Parts of Congruent Triangles), we have \( \angle OXM = \angle OXN \).
- Therefore, the line segment \( OX \) bisects the angle \( MXN \).

#### Case (ii): Chords intersect on the circle

1. **Draw the Circle and Chords**:
- Draw a circle and label its center as \( O \).
- Draw two equal chords \( AB \) and \( CD \) that intersect at point \( X \) on the circle.

2. **Draw Perpendiculars from the Center**:
- From point \( O \), draw perpendiculars to the chords \( AB \) and \( CD \) at points \( M \) and \( N \) respectively.

3. **Identify Triangles**:
- Consider triangles \( \triangle OXM \) and \( \triangle OXN \).

4. **Establish Congruence**:
- In triangles \( OXM \) and \( OXN \):
- \( OX = OX \) (common side),
- \( \angle OMX = \angle ONX = 90^\circ \) (perpendiculars),
- \( OM = ON \) (equal distance from center to equal chords).
- By RHS congruence, \( \triangle OXM \cong \triangle OXN \).

5. **Use CPCT**:
- By CPCT, we have \( \angle MXO = \angle NXO \).
- Therefore, the line segment \( OX \) bisects the angle \( MXN \).

#### Case (iii): Chords intersect outside the circle

1. **Draw the Circle and Chords**:
- Draw a circle and label its center as \( O \).
- Draw two equal chords \( AB \) and \( CD \) that intersect at point \( X \) outside the circle.

2. **Draw Perpendiculars from the Center**:
- From point \( O \), draw perpendiculars to the extended lines of chords \( AB \) and \( CD \) at points \( M \) and \( N \) respectively.

3. **Identify Triangles**:
- Consider triangles \( \triangle OXM \) and \( \triangle OXN \).

4. **Establish Congruence**:
- In triangles \( OXM \) and \( OXN \):
- \( OX = OX \) (common side),
- \( \angle XMO = \angle XNO = 90^\circ \) (perpendiculars),
- \( OM = ON \) (equal distance from center to equal chords).
- By RHS congruence, \( \triangle OXM \cong \triangle OXN \).

5. **Use CPCT**:
- By CPCT, we have \( \angle MXO = \angle NXO \).
- Therefore, the line segment \( OX \) bisects the angle \( MXN \).

### Conclusion:
In all three cases, we have shown that the line segment joining the point of intersection of the chords to the center of the circle bisects the angle between the chords.