ESTION 2

To prove that if \( OM \perp AB \) (where \( O \) is the center of the circle, \( A \) and \( B \) are points on the circumference, and \( M \) is a point on chord \( AB \)), then \( AM = MB \), we can follow these steps:

### Step-by-step proof:

1. **Construct Radius Lines**:
Draw the radii \( OA \) and \( OB \). Since \( A \) and \( B \) lie on the circumference, both lines are equal in length (as they are radii of the same circle):
\[
OA = OB = r \quad (where \, r \, is \, the \, radius \, of \, the \, circle)
\]

2. **Right Triangle Formed**:
Since \( OM \perp AB \), we have two right triangles: \( OMA \) and \( OMB \).

3. **Use Right Triangle Properties**:
In triangles \( OMA \) and \( OMB \):
- Both triangles share the side \( OM \).
- \( OA \) and \( OB \) are both radii of the circle and thus equal.

4. **Applying the Pythagorean Theorem**:
According to the Pythagorean theorem:
\[
OA^2 = OM^2 + AM^2 \quad \text{and} \quad OB^2 = OM^2 + MB^2
\]
Substituting \( OA = OB \):
\[
OA^2 = OB^2 \implies OM^2 + AM^2 = OM^2 + MB^2
\]

5. **Simplify the Equation**:
By subtracting \( OM^2 \) from both sides, we arrive at:
\[
AM^2 = MB^2
\]

6. **Conclusion**:
Taking the square root of both sides gives:
\[
AM = MB
\]
Therefore, we have proven that if \( OM \perp AB \), then \( AM = MB \).

### Diagram:

Here's a simple representation of the diagram you can sketch to illustrate the proof:

```
B
/ \
/ \
/ \
O------M
\ /
\ /
\ /
A
```

- The point \( O \) is the center of the circle.
- \( A \) and \( B \) are points on the circumference.
- \( M \) is the midpoint of chord \( AB \), where \( OM \perp AB \).
- The line segments \( OA \) and \( OB \) are equal, and \( AM \) is equal to \( MB \).

### Summary:
Thus, the proof demonstrates that if \( OM \) is perpendicular to chord \( AB \), the segment \( M \) divides the chord \( AB \) into two equal parts, establishing \( AM = MB \).