Equidistant Chords Theorem

This theorem will be proven in two parts.

Part 1

Consider a circle with center E. Also, consider two chords AB and CD equidistant from the center. Recall that the distance between a point and a chord is the length of the perpendicular segment from the point to the chord. Therefore, EF and AB are perpendicular, and EG and CD are also perpendicular.

Furthermore, EF is a part of a radius and, consequently, of a diameter. By the Perpendicular Chord Bisector Theorem, EF bisects the chord AB. Similarly, EG bisects the chord CD. AB=2AF andCD=2CG This can be visualized in the diagram.

In a circle, all radii are congruent. Therefore, EA and EC are congruent segments.

Note that △ AEF and △ CEG are right triangles. By the Hypotenuse Leg Theorem, these triangles are congruent. Since corresponding parts of congruent figures are congruent, it can be stated that AF and CG are congruent. △ AEF ≅ △ CEG ⇓ AF ≅ CG By definition, congruent segments have the same length. Therefore, AF=CG. This can be substituted into the equation AB=2AF. Recall that 2CG=CD. Therefore, chords AB and CD are congruent.

It has been proven that if two chords are equidistant from the center of the circle, then they are congruent.

Part 2

Consider a circle centered at point E with two congruent chords.

Now, consider two radii that are perpendicular to the chords AB and CD. Let F and G be the points of intersection.

Note that EF, which is part of a radius and a diameter, is perpendicular to AB. Therefore, by the Perpendicular Chord Bisector Theorem, EF bisects AB. Similarly, EG bisects CD. Additionally, since AB and CD are congruent, AF, FB, CG, and GD are equal.

Since all radii of a circle are congruent, it can be said that EA and EC are congruent segments. Therefore, △ AEF and △ CEG are right triangles with one pair of congruent legs and congruent hypotenuses.

By the Hypotenuse Leg Theorem, it can be concluded that △ AEF and △ CEG are congruent triangles. Since corresponding parts of congruent figures are congruent, EF and EG are congruent. This means that the segments have the same length. Note that these are the distances from the chords AB and CD to the center of the circle.

It has been proven that if two chords are congruent, then they are equidistant from the center of the circle.