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.
In the words of Euclid:
- Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the center is always greater than the more remote.
(The Elements: Book $\text{III}$: Proposition $15$)
Proof
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.
$\Box$
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.
$\blacksquare$
Historical Note
This proof is Proposition $15$ of Book $\text{III}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions