butterfly.md

The butterfly theorem can be stated as: Let M be the midpoint of a chord AB of a circle, through which two other chords PQ and RS are drawn; PS and QR intersect chord AB at C and D correspondingly. Then M is the midpoint of CD. ^[1]

This theorem can be generalized in two forms:

1. Generalize the circle to any conic section;
2. Generalize the condition MA = MB and MC = MD to or .

Proof by Analytic Geometry

Here we try to prove the combination of 2 generalized forms. ^[2]

We put M onto the origin of Cartesian coordinates and B onto positive x-axis.

Step 1

The conic APRBQS can be represented as:

(Let's assume M not on the conic so that f ≠ 0.)

The coordinates of A and B are:

Make y = 0 then we get . So x_A and x_B are two roots of this equation.

Then we get:

Step 2a

The coordinates of C and D can be calculated by:

Let's assume that neither PQ nor RS is perpendicular to AB, then PQ and RS can be represented as y = qx and y = rx. (We also assume PQ and RS not coinciding so that q ≠ r.)

(The case either PQ or RS perpendicular to AB will be proved in Step 2b.)

Then all y-coordinates can be replaced with 0 or x-coordinates:

We solve x_C and x_D as:

(Here x_C and x_D should exist because y_P ≠ y_S and y_Q ≠ y_R.)

Then we get:

Step 3a

Note that x_P and x_Q are two roots of equations:

So we have:

where . (Here we don't consider Q = 0 because the line intersects the conic at two different points.)

Analogously, we have:

where .

Finally, we get: ^[3]

Step 2b

If RS is perpendicular to AB, we can still use Eq. 2a but keep y_R and y_S not replaced:

We solve x_C and x_D as:

(Here x_C and x_D should exist because y_P ≠ y_S and y_Q ≠ y_R.)

Then we get:

Step 3b

Note that y_R and y_S are two roots of equations:

So we have:

And we still have Eq. 3a.

Finally, we get:

Degenerate Conic

The generalized butterfly theorem still holds if the conic is degenerated to two lines.

In the below special case, AB is a chord of two lines ASQ and BPR, and M is a point on AB, through which two other lines MPQ and MSR are drawn; PS and QR intersect chord AB at the same point C (lines AB, PS and QR are concurrent at point C).

According to the generalized butterfly theorem:

In other words, (M, C) and (A, B) are harmonic conjugates: ^[4]

Notes

1. Here we use the diagram from Mathoman's website. The diagram from Wikipedia is wrong because M is not the midpoint of PQ and XY.
2. A similar proof (but not generalized forms) can be found here (Proof 18).
3. Here is a proof by SymPy without Vieta's formulas.
4. A simple proof of projective harmonic conjugate can be found here (Problem 1).