Consider chord $AB$ in the given circle with midpoint $M$. If $CD$ and $EF$ are two more chords passing through $M$ such that $CF$ and $ED$ meets $AB$ at $G$ and $H$ respectively. Then $M$ is also the midpoint of $GH$.
For our convenience, let $AM = BM =a$, $GM=b$ and $HM=c$. Drop perpendiculars $GG'$, $GG''$, $HH'$ and $HH''$ on $CM, FM, EM$ and $DM$ respectively (and let their lengths be $b',b'',c'$ and $c''$ resp.). We have shown the construction below:
We now chase the similar triangles. Since $\triangle MG'G \sim \triangle MH''H$ we have $\frac{b}{c} = \frac{b'}{c''}$. Similarly $ \frac{b}{c} = \frac{b''}{c'}$. Also note $\triangle CG'G \sim \triangle EH'H$ and $\triangle FG''G \sim \triangle DH''H$. So $\frac{b'}{c'} = \frac{CG}{EH}$ and $\frac{b''}{c''} = \frac{FG}{DH}$. So we have, $$\frac{b^2}{c^2}=\frac{b'b''}{c'c''}=\frac{CG \times FG}{DH \times EH}$$Intersecting chords theorem gives us $CG \times FG = AG \times BG=(a-b)(a+b)$ and similarly $DH \times EH = (a-c)(a+c)$. Thus we have $\frac{b^2}{c^2}= \frac{a^2-b^2}{a^2-c^2}$ which easily gives us $b=c$.
References:
- A Sequel to the First Six Books of Euclid by John Casey
- David Joyce's web version of Euclid's Elements
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