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Let's go through the problem step by step.

### Part (a): Prove that \( \triangle AOM \equiv \triangle BOM \) using SSS (Side-Side-Side Congruence).

We need to show that the three sides of \( \triangle AOM \) and \( \triangle BOM \) are equal to apply the **SSS congruence** rule.

1. **Side 1: \( OA = OB \)**
   - Since \( O \) is the center of the circle, both \( A \) and \( B \) are points on the circle, so \( OA \) and \( OB \) are **radii** of the circle.
   - Hence, \( OA = OB \).

2. **Side 2: \( AM = MB \)**
   - The point \( M \) is given as the **midpoint** of the chord \( AB \), which means \( AM = MB \).

3. **Side 3: \( OM \) is common**
   - The line segment \( OM \) is common to both triangles \( \triangle AOM \) and \( \triangle BOM \).

Since all three pairs of corresponding sides are equal, we can conclude that:
\[
\triangle AOM \equiv \triangle BOM \quad \text{(SSS criterion)}
\]

### Part (b): Prove that the line drawn from the center of a circle to the midpoint of a chord is perpendicular to the chord.

Using the result from part (a), since \( \triangle AOM \equiv \triangle BOM \), it follows that:

1. **Corresponding Angles**: The angles \( \angle AMO \) and \( \angle BMO \) are **equal**.
2. **Linear Pair**: Together, these angles form a linear pair, meaning \( \angle AMO + \angle BMO = 180^\circ \).
3. **Equality**: Since \( \angle AMO = \angle BMO \), the only way for their sum to be \( 180^\circ \) is if \( \angle AMO = \angle BMO = 90^\circ \).

Thus, \( OM \), the line from the center \( O \) to the midpoint \( M \), is perpendicular to the chord \( AB \).

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Would you like more details or have any further questions?

Here are 5 questions related to this problem:
1. What is the importance of using the midpoint in this type of geometry problem?
2. Can you explain why the radii of a circle are always equal?
3. What other methods could be used to prove that two triangles are congruent besides SSS?
4. How can we generalize this result for any chord of a circle?
5. What is the relationship between the perpendicular bisector of a chord and the center of a circle?

**Tip**: In geometry, when a line bisects a chord and is drawn from the center, it is always perpendicular to the chord.

Prove that ΔAOM ≡ ΔBOM using SSS. Hence, prove that the line drawn from the center of a circle to the midpoint of a chord is perpendicular to the chord.

Learn how to prove that triangles AOM and BOM are congruent using the SSS rule, and that the line from the center of a circle to the midpoint of a chord is perpendicular. This problem is ideal for students in Grades 9-10 and covers essential circle geometry concepts.

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