Prove the theorem that states that if 𝑂𝑀 ⊥ 𝐴𝐵 then 𝐴𝑀 = 𝑀𝐵. Redraw the diagram above to show any necessary constructions.

To prove the theorem that states "If \( OM \perp AB \), then \( AM = MB \)", we will use basic properties of circles and perpendicular bisectors. Here's a step-by-step proof, along with the necessary construction.

### Proof Outline:

#### Given:
- \( O \) is the center of the circle.
- \( AB \) is a chord of the circle.
- \( M \) is the point where \( OM \) intersects \( AB \), with \( OM \) being perpendicular to \( AB \).

#### To Prove:
- \( AM = MB \) (i.e., \( M \) is the midpoint of \( AB \)).

#### Construction:
- Draw the radius \( OA \) and \( OB \) from the center \( O \) to the points \( A \) and \( B \) respectively.
- Draw the perpendicular line \( OM \) from the center \( O \) to the chord \( AB \).

#### Diagram:

I'll describe how you can draw the diagram:

1. **Circle**: Start by drawing a circle with center \( O \).
2. **Chord \( AB \)**: Draw a chord \( AB \) in the circle.
3. **Line \( OM \)**: Draw a line from \( O \) perpendicular to \( AB \), intersecting \( AB \) at point \( M \).
4. **Radii \( OA \) and \( OB \)**: Draw radii \( OA \) and \( OB \).

#### Proof:
1. **Right Triangles**:
   - Since \( OM \) is perpendicular to \( AB \), the triangles \( \triangle OMA \) and \( \triangle OMB \) are right triangles.
   
2. **Radii Equality**:
   - \( OA = OB \) because both are radii of the circle.
   
3. **Shared Length**:
   - \( OM \) is common to both triangles \( \triangle OMA \) and \( \triangle OMB \).
   
4. **Congruence of Triangles**:
   - By the Hypotenuse-Leg (HL) theorem, which states that two right triangles are congruent if their hypotenuse and one leg are equal, \( \triangle OMA \cong \triangle OMB \).
   
5. **Corresponding Parts of Congruent Triangles**:
   - Since \( \triangle OMA \cong \triangle OMB \), corresponding sides are equal. Therefore, \( AM = MB \).

This completes the proof that \( M \) is the midpoint of \( AB \), meaning \( AM = MB \).

### Follow-up Questions:
1. How would this proof change if \( OM \) was not perpendicular to \( AB \)?
2. What if \( AB \) is a diameter? What can be said about \( M \) in this case?
3. Can this theorem be extended to chords that are not necessarily within the same circle? Why or why not?
4. How does the perpendicular bisector theorem apply to this proof?
5. What role does symmetry play in the proof of this theorem?

### Tip:
In circle geometry, the perpendicular from the center of a circle to a chord always bisects the chord. This is a useful property when solving problems involving chords and radii.

Learn how to prove the theorem that states "If OM is perpendicular to AB, then AM equals MB" with a detailed step-by-step proof. This problem covers circle geometry, perpendicular bisectors, and congruent triangles. Suitable for high school students.

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