Bunch of Thoughts: July 2024

Saturday, 20 July 2024

This is a simple geometric result, first published in early 18th century.

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:

Bunch of Thoughts