# Congruent Chords in the Same Circle are Equidistant from the Center

Two congruent chords of a circle are equidistant from its center.
NOTE: Two chords are said to be congruent if they are equal in length.

Consider the equation of the circle

Suppose that $AB$ and $CD$ are congruent chords with $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$, $C\left( {{x_3},{y_3}} \right)$ and $D\left( {{x_4},{y_4}} \right)$ as shown in the given diagram. Since the circle passes through the points $A,\,B,\,C$ and $D$, the equation of the circle becomes

Since $M$ is the midpoint of the chord $AB$, so

Since $N$ is the midpoint of the chord $CD$, so

Now we shall find the distance between $O$ and $M$, as follows:

Similarly, we can show that

Since $AB$ and $CD$ are congruent chords,  ${\left| {AB} \right|^2} = {\left| {CD} \right|^2}$

Using equations (2) and (3), we get the following result:

Adding ${r^2}$ to both sides of the above equation, we have

Using equations (vi) and (vii) in equation (viii), we get

This shows that the congruent chords of a circle are equidistant from its center.