Butterfly theorem

The butterfly theorem is a classical result in Euclidean geometry, which can be stated as follows:^{[1] : p. 78}

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.

Since

${\displaystyle \triangle MXX'\sim \triangle MYY',}${\displaystyle \triangle MXX'\sim \triangle MYY',}

${\displaystyle {MX \over MY}={XX' \over YY'},}${\displaystyle {MX \over MY}={XX' \over YY'},}

${\displaystyle \triangle MXX''\sim \triangle MYY'',}${\displaystyle \triangle MXX''\sim \triangle MYY'',}

${\displaystyle {MX \over MY}={XX'' \over YY''},}${\displaystyle {MX \over MY}={XX'' \over YY''},}

${\displaystyle \triangle AXX'\sim \triangle CYY'',}${\displaystyle \triangle AXX'\sim \triangle CYY'',}

${\displaystyle {XX' \over YY''}={AX \over CY},}${\displaystyle {XX' \over YY''}={AX \over CY},}

${\displaystyle \triangle DXX''\sim \triangle BYY',}${\displaystyle \triangle DXX''\sim \triangle BYY',}

${\displaystyle {XX'' \over YY'}={DX \over BY}.}${\displaystyle {XX'' \over YY'}={DX \over BY}.}

From the preceding equations and the intersecting chords theorem, it can be seen that

${\displaystyle \left({MX \over MY}\right)^{2}={XX' \over YY'}{XX'' \over YY''},}${\displaystyle \left({MX \over MY}\right)^{2}={XX' \over YY'}{XX'' \over YY''},}

${\displaystyle {}={AX\cdot DX \over CY\cdot BY},}${\displaystyle {}={AX\cdot DX \over CY\cdot BY},}

${\displaystyle {}={PX\cdot QX \over PY\cdot QY},}${\displaystyle {}={PX\cdot QX \over PY\cdot QY},}

${\displaystyle {}={(PM-XM)\cdot (MQ+XM) \over (PM+MY)\cdot (QM-MY)},}${\displaystyle {}={(PM-XM)\cdot (MQ+XM) \over (PM+MY)\cdot (QM-MY)},}

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

since PM = MQ.

So,

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

Cross-multiplying in the latter equation,

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

Cancelling the common term

${\displaystyle {-(MX)^{2}\cdot (MY)^{2}}}${\displaystyle {-(MX)^{2}\cdot (MY)^{2}}}

from both sides of the equation yields

${\displaystyle {(MX)^{2}\cdot (PM)^{2}}={(MY)^{2}\cdot (PM)^{2}},}${\displaystyle {(MX)^{2}\cdot (PM)^{2}}={(MY)^{2}\cdot (PM)^{2}},}

hence MX = MY, since MX, MY, and PM are all positive, real numbers.

Thus, M is the midpoint of XY.

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. Reverend Thomas Scurr asked the same question again in 1814 in the Gentleman's Diary or Mathematical Repository.^[4]

• The Butterfly Theorem at cut-the-knot
• A Better Butterfly Theorem at cut-the-knot
• Proof of Butterfly Theorem at PlanetMath
• The Butterfly Theorem by Jay Warendorff, the Wolfram Demonstrations Project.
• Weisstein, Eric W. "Butterfly Theorem". MathWorld .

• Euclidean plane geometry
• Theorems about circles

• Articles with short description
• Short description is different from Wikidata
• Commons category link is on Wikidata
• Articles containing proofs