Out of two unequal chords of a circle, the bigger chord is closer to the centre of the circle. Prove it.

Answer

To prove that out of two unequal chords of a circle, the bigger chord is closer to the center of the circle, we can follow these steps:

### Step 1: Draw the Circle and Chords
Draw a circle with center O. Label the two unequal chords as AB (the bigger chord) and CD (the smaller chord).

**Hint:** Visualizing the problem with a diagram can help clarify the relationships between the elements involved.

### Step 2: Draw Perpendiculars from the Center
Draw perpendiculars from the center O to both chords AB and CD. Let the foot of the perpendicular from O to AB be M and from O to CD be N.

**Hint:** Remember that the perpendicular from the center of a circle to a chord bisects the chord.

### Step 3: Identify the Lengths
Let the lengths of the chords be:
- AB = 2 * MB (since M is the midpoint of AB)
- CD = 2 * ND (since N is the midpoint of CD)

### Step 4: Apply the Pythagorean Theorem
In triangle OMB, apply the Pythagorean theorem:
\[ OB^2 = OM^2 + MB^2 \]
Since OB is the radius (R), we have:
\[ R^2 = OM^2 + MB^2 \]
From this, we can express MB:
\[ MB = \frac{AB}{2} \]

**Hint:** The relationship between the radius and the segments created by the perpendicular can be useful for calculations.

### Step 5: Write the Equation for Chord AB
Substituting MB into the equation gives:
\[ R^2 = OM^2 + \left(\frac{AB}{2}\right)^2 \]
Rearranging yields:
\[ AB^2 = 4(R^2 - OM^2) \]
Let this be Equation (1).

### Step 6: Apply the Pythagorean Theorem to the Smaller Chord
Now, apply the Pythagorean theorem in triangle OND:
\[ OD^2 = ON^2 + ND^2 \]
Again, since OD is also the radius (R), we have:
\[ R^2 = ON^2 + ND^2 \]
And similarly, express ND:
\[ ND = \frac{CD}{2} \]

### Step 7: Write the Equation for Chord CD
Substituting ND into the equation gives:
\[ R^2 = ON^2 + \left(\frac{CD}{2}\right)^2 \]
Rearranging yields:
\[ CD^2 = 4(R^2 - ON^2) \]
Let this be Equation (2).

### Step 8: Compare the Two Chords
Since we know that AB > CD, we can write:
\[ AB^2 > CD^2 \]
Substituting from Equations (1) and (2):
\[ 4(R^2 - OM^2) > 4(R^2 - ON^2) \]

### Step 9: Simplify the Inequality
Dividing both sides by 4 gives:
\[ R^2 - OM^2 > R^2 - ON^2 \]
This simplifies to:
\[ -OM^2 > -ON^2 \]
Multiplying through by -1 (which reverses the inequality):
\[ OM^2 < ON^2 \]

### Step 10: Conclude the Proof
Taking the square root of both sides (since lengths are positive):
\[ OM < ON \]
Thus, we have proven that the distance from the center O to the bigger chord AB (OM) is less than the distance from the center O to the smaller chord CD (ON).

**Final Conclusion:** The bigger chord is indeed closer to the center of the circle.

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