The butterfly theorem is a classical result in Euclidean geometry, which can be stated as follows:[1]
Let be the midpoint of a chord of a circle, through which two other chords and are drawn; and intersect chord at and correspondingly. Then is the midpoint of .
A formal proof of the theorem is as follows:Let the perpendiculars and be dropped from the point on the straight lines and respectively. Similarly, let and be dropped from the point perpendicular to the straight lines and respectively.
Since
\triangleMXX'\sim\triangleMYY',
{MX\overMY}={XX'\overYY'},
\triangleMXX''\sim\triangleMYY'',
{MX\overMY}={XX''\overYY''},
\triangleAXX'\sim\triangleCYY'',
{XX'\overYY''}={AX\overCY},
\triangleDXX''\sim\triangleBYY',
{XX''\overYY'}={DX\overBY}.
From the preceding equations and the intersecting chords theorem, it can be seen that
\left({MX\overMY}\right)2={XX'\overYY'}{XX''\overYY''},
{}={AX ⋅ DX\overCY ⋅ BY},
{}={PX ⋅ QX\overPY ⋅ QY},
{}={(PM-XM) ⋅ (MQ+XM)\over(PM+MY) ⋅ (QM-MY)},
{}={(PM)2-(MX)2\over(PM)2-(MY)2},
since .
So
{(MX)2\over(MY)2}={(PM)2-(MX)2\over(PM)2-(MY)2}.
Cross-multiplying in the latter equation,
{(MX)2 ⋅ (PM)2-(MX)2 ⋅ (MY)2}={(MY)2 ⋅ (PM)2-(MX)2 ⋅ (MY)2}.
Cancelling the common term
{-(MX)2 ⋅ (MY)2}
from both sides of the resulting equation yields
{(MX)2 ⋅ (PM)2}={(MY)2 ⋅ (PM)2},
hence, since MX, MY, and PM are all positive, real numbers.
Thus, is the midpoint of .
Other proofs exist,[2] including one using projective geometry.[3]
Proving the butterfly theorem was posed as a problem by William Wallace in The Gentleman's Mathematical Companion (1803). Three solutions were published in 1804, and in 1805 Sir William Herschel posed the question again in a letter to Wallace. Rev. Thomas Scurr asked the same question again in 1814 in the Gentleman's Diary or Mathematical Repository.[4]