Butterfly theorem

For the "butterfly lemma" of group theory, see Zassenhaus lemma.

The butterfly theorem is a classical result in Euclidean geometry, which can be stated as follows:

Let M be the midpoint of a chord PQ of a circle, through which two other chords AB and CD are drawn; AD and BC intersect chord PQ at X and Y correspondingly. Then M is the midpoint of XY.

A formal proof of the theorem is as follows: Let the perpendiculars $XX'\,$ and $XX''\,$ be dropped from the point $X\,$ on the straight lines $AM\,$ and $DM\,$ respectively. Similarly, let $YY'\,$ and $YY''\,$ be dropped from the point $Y\,$ perpendicular to the straight lines $BM\,$ and $CM\,$ respectively.

Proof of Butterfly theorem

Now, since

$\triangle MXX' \sim \triangle MYY',\,$
${MX \over MY} = {XX' \over YY'},$
$\triangle MXX'' \sim \triangle MYY'',\,$
${MX \over MY} = {XX'' \over YY''},$
$\triangle AXX' \sim \triangle CYY'',\,$
${XX' \over YY''} = {AX \over CY},$
$\triangle DXX'' \sim \triangle BYY',\,$
${XX'' \over YY'} = {DX \over BY},$

From the preceding equations, it can be easily seen that

$\left({MX \over MY}\right)^2 = {XX' \over YY' } {XX'' \over YY''},$
${} = {AX.DX \over CY.BY},$
${} = {PX.QX \over PY.QY},$
${} = {(PM-XM).(MQ+XM) \over (PM+MY).(QM-MY)},$
${} = { (PM)^2 - (MX)^2 \over (PM)^2 - (MY)^2},$

since $PM \,$ = $MQ \,$

Now,

${ (MX)^2 \over (MY)^2} = {(PM)^2 - (MX)^2 \over (PM)^2 - (MY)^2}.$

So, it can be concluded that $MX = MY, \,$ or $M \,$ is the midpoint of $XY. \,$

An alternate proof using projective geometry can be found in problem 8 of the link below.

http://www.imomath.com/index.php?options=628&lmm=0

Bibliography

H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967.