Prove that equal chords of a circle subtend equal angles at the centre.

Answer

To prove that equal chords of a circle subtend equal angles at the center, we can follow these steps:

### Step 1: Draw the Circle and Chords
Let’s consider a circle with center O. Draw two equal chords AB and CD such that AB = CD.

### Step 2: Identify the Triangles
We will analyze the triangles formed by the center O and the endpoints of the chords. We have two triangles: triangle AOB and triangle COD.

### Step 3: State the Given Information
- Given: Chords AB and CD are equal, i.e., AB = CD.
- OA and OC are radii of the circle.
- OB and OD are also radii of the circle.

### Step 4: Establish Equalities
From the above information, we can establish the following equalities:
1. AB = CD (Given)
2. OA = OC (Both are radii of the circle)
3. OB = OD (Both are radii of the circle)

### Step 5: Apply the SSS Congruence Criterion
Now we have:
- AB = CD
- OA = OC
- OB = OD

Since we have three pairs of equal sides, we can conclude that triangle AOB is congruent to triangle COD by the Side-Side-Side (SSS) congruence criterion.

### Step 6: Conclude the Angles
Since the triangles AOB and COD are congruent, their corresponding angles are equal. Thus, we have:
∠AOB = ∠COD (by CPCT, Corresponding Parts of Congruent Triangles).

### Final Conclusion
Therefore, we have proved that equal chords of a circle subtend equal angles at the center.

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