Chords that have an equal length are called congruent. An interesting property of such chords is that regardless of their position in the circle, they are all an equal distance from the circle’s center. The distance is defined as the length of a perpendicular line from a point to a line.

## Problem

In circle O, the two chords AB and CD are congruent. Prove that they are equidistant from the point O.

## Strategy

We can use triangle congruency to prove line segments are equal length. Since the distance is defined as the length of a perpendicular line from a point to a line, we are dealing with a perpendicular line to the chord, and we can rely on theorems about chords. Specifically, that a diameter or a radius that is perpendicular to a chord bisects that chord.

Since OE and OF are such lines that are perpendicular to the chords, they are also bisectors, and if we construct triangles using radii, as below, it is easy to show that they are congruent using the Hypotenuse-Leg postulate.

## Proof

(1) AB=CD //Given, defintion of congruent chords

(2) OE⊥AB //Given, Defintion of distance

(3) OF⊥CD //Given, Defintion of distance

(4) AE=EB=½AB //line from center of circle and perpendicular to a chord bisects the chord

(5) CF=FD=½CD //line from center of circle and perpendicular to a chord bisects the chord

(6) AE = CF //(1), (4), (5), halves of equal segments, transitive property of equality

(7) m∠OEA = m∠OFC=90° //(2), (3) defintion of perpendicular lines

(8) ΔOAE ≅ ΔOCF //Hypotenuse-Leg

(9) OE=OF //Corresponding sides in congruent triangles (CPCTC)