The Diameter is Longer Than Any Chord That Doesn't Pass Through the Center

In a circle, the diameter is always longer than any chord that doesn't pass through the center.

Proof

Consider a circle with center O, a chord AB that doesn't pass through the center, and a diameter CD.

The points O, A, and B form a right triangle OAB.

According to the triangle inequality theorem, in a triangle, each side is shorter than the sum of the other two sides.

Therefore, chord AB is shorter than the sum of OA and OB.

$$ \overline{AB} < \overline{OA} + \overline{OB} $$

Both OA and OB are equal to the radius of the circle, so OA=r and OB=r.

Thus, we can rewrite the inequality by substituting OA and OB with the radius (r) of the circle.

$$ \overline{AB} < r + r $$

$$ \overline{AB} < 2r $$

Knowing that segment CD is the diameter (d) of the circle and equals the sum of two radii, CD=2r, we can replace 2r with the diameter (d).

$$ \overline{AB} < 2r $$

Since the diameter d=2r is twice the radius,

$$ \overline{AB} < d $$

This proves that any chord that doesn't pass through the center of the circle is shorter than the diameter (d).

Alternative Proof

Consider a circle with center O, a chord AB that doesn't pass through the center, and a diameter CD.

Draw the segment OB.

Draw a radius that passes through the midpoint E of chord AB.

Knowing that the perpendicular bisector of a chord always passes through the center of the circle, we deduce that a radius that bisects a chord is perpendicular to the chord itself.

 

Therefore, triangle OBE is a right triangle.

According to the Pythagorean theorem, in a right triangle, the hypotenuse is always longer than any of the legs. Thus, OB>BE.

$$ \overline{OB} > \overline{BE} $$

By the invariant property, we multiply both sides of the inequality by two.

$$ 2 \cdot \overline{OB} > 2 \cdot \overline{BE} $$

Knowing that OB=r is equal to the radius, and that twice the radius 2r=CD equals the diameter (d),

$$ \overline{CD} > 2 \cdot \overline{BE} $$

Since E is the midpoint of chord AB, 2BE is equal to chord AB.

$$ \overline{CD} > \overline{AB} $$

This proves that the diameter (CD) is always longer than any chord (AB) that doesn't pass through the center of the circle.

And so on.