This theorem will be proven in two parts.

Part $1$

Consider a circle with center $E.$ Also, consider two chords $\overline{AB}$ and $\overline{CD}$ 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, $\overline{EF}$ and $\overline{AB}$ are perpendicular, and $\overline{EG}$ and $\overline{CD}$ are also perpendicular.

Furthermore, $\overline{EF}$ is a part of a radius and, consequently, of a diameter. By the Perpendicular Chord Bisector Theorem, $\overline{EF}$ bisects the chord $\overline{AB}.$ Similarly, $\overline{EG}$ bisects the chord $\overline{CD}.$ This can be visualized in the diagram.

In a circle, all radii are congruent. Therefore, $\overline{EA}$ and $\overline{EC}$ 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 $\overline{AF}$ and $\overline{CG}$ are congruent. 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 $\overline{AB}$ and $\overline{CD}$ 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 $\overline{AB}$ and $\overline{CD}.$ Let $F$ and $G$ be the points of intersection.

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

Since all radii of a circle are congruent, it can be said that $\overline{EA}$ and $\overline{EC}$ 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, $\overline{EF}$ and $\overline{EG}$ are congruent. This means that the segments have the same length. Note that these are the distances from the chords $\overline{AB}$ and $\overline{CD}$ 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.