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:
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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): \[ OA = OB = r \quad (where , r , is , the , radius , of , the , circle) \]
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Right Triangle Formed: Since \( OM \perp AB \), we have two right triangles: \( OMA \) and \( OMB \).
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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.
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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 \( OA = OB \): \[ OA^2 = OB^2 \implies OM^2 + AM^2 = OM^2 + MB^2 \]
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Simplify the Equation: By subtracting \( OM^2 \) from both sides, we arrive at: \[ AM^2 = MB^2 \]
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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
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O------M
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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 \( AM = MB \).
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