Equal Chords in Circle

Theorem
In a circle, equal chords are equally distant from the center, and chords that are equally distant from the center are equal in length.

Proof

 * Euclid-III-14.png

Let $ABDC$ be a circle, and let $AB$ and $CD$ be equal chords on it.

From Finding Center of Circle, let $E$ be the center of $ABDC$.

Construct $EF$, $EG$ perpendicular to $AB$ and $CD$, and join $AE$ and $EC$.

From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AD$.

So $AF = FB$ and so $AB = 2 AF$.

Similarly $CD = 2CG$.

As $AB = CD$ it follows that $AE = CG$.

Since $AE = EC$ the square on $AE$ equals the square on $EC$.

We have that $\angle AFE$ is a right angle.

So from Pythagoras's Theorem the square on $AE$ equals the sum of the squares on $AF$ and $FE$.

Similarly, the square on $EC$ equals the sum of the squares on $EG$ and $GC$.

Thus the square on $EF$ equals the square on $EG$.

So $EF = EG$.

So from Definition III: 4, chords are at equal distance from the center when the perpendiculars drawn to them from the center are equal.

So $AB$ and $CD$ are at equal distance from the center.

Now suppose that $AB$ and $CD$ are at equal distance from the center.

That is, let $EF = EG$.

Using the same construction as above, it is proved similarly that $AB = 2AF$ and $CD = 2CG$.

Since $AE = CE$, the square on $AE$ equals the square on $CE$.

From Pythagoras's Theorem:
 * the square on $AE$ equals the sum of the squares on $AF$ and $FE$
 * the square on $EC$ equals the sum of the squares on $EG$ and $GC$.

So the sum of the squares on $AF$ and $FE$ equals the sum of the squares on $EG$ and $GC$.

But the square on $FE$ equals the square on $EG$.

So the square on $AF$ equals the square on $CG$, and so $AF = CG$.

As $AB = 2AF$ and $CD = 2CG$ it follows that $AB = CD$.

Hence the result.