Relative Lengths of Chords of Circles

Theorem
Of chords in a circle, the diameter is the greatest, and of the rest the nearer to the center is always greater than the more remote.

Proof

 * Euclid-III-15.png

Let $ABCD$ be a clrcle, let $AD$ be its diameter and $E$ the center.

Let $BC$ and $FG$ be chords of $ABCD$, where $BC$ is nearer to the center than $FG$.

Let $EH$ and $EK$ be drawn perpendicular to $BC$ and $FG$ respectively.

Because $BC$ is nearer to the center than $FG$, it follows from Book III: Definition 5 that $EK$ is longer than $EH$.

Position $L$ on $EK$ so that $EL = EH$, and draw $MN$ through $L$ perpendicular to $FG$.

Join $EM$, $EN$, $EF$ and $EG$.

Since $EH = EL$ it follows from Equal Chords in Circle that $BC = MN$.

Also, since $AE = EM$ and $ED = EN$, $AD = ME + EN$.

But $ME + EN > MN$ from Sum of Two Sides of Triangle Greater than Third Side, and $MN = BC$.

So $AD > BC$.

Thus we have that the diameter is greater than any other chord of a circle.

Since $ME = FE$ and $EN = EG$ and $\angle MEN > \angle FEG$, from the Hinge Theorem it follows that $MN > FG$.

But $MN = BC$.

So a chord nearer the center is greater than one more remote.