## Algebraic proof of the theorem of butterflies in quadrilaterals

### Fan Tai-Sheng

February 5, 2011

Republic of China (Taiwan)

Through the intersection I of the diagonals AC,BD of a quadrilateral ABCD, draw two lines EF and HG that meet the sides of ABCD in E, F, G, H. Let M and N be the intersections of EG and FH with AC. Then

1/IM - 1/IA =1/IN - 1/IC.

In the figure, in Cartesian coordinates, let AC lie on the x-axis, with I being the origin, and points A,B,C,D be

AB: b(x + a) + (e - a)y = 0

CD: bd(x - c) + (de - c)y = 0

BC: b(x - c) + (c + e)y = 0

AD: bd(x + a) + (de + a)y = 0.

Since the lines EF and GH both pass through the origin I, they can be given by equations

EF: y - m_{1}x = 0 and

GH: y - m_{2}x = 0,

where m_{1} and m_{2} are the slopes of EF and GH respectively. Extend EF to cut the sides BC and AD at H'
and G' respectively and also extend GH to cut the sides AB and CD at E' and F' respectively.

Now it can be easily seen that for parameters λ and μ, the equation

f(x,y) | = λAB(x,y)CD(x,y)BC(x,y)AD(x,y) + μ[EF(x,y)]²[GH(x,y)]² | |

= λ[b(x + a) + (e - a)y][bd(x - c) + (de - c)y][b(x - c) + (c + e)y][bd(x + a) + (de + a)y] + μ(y - m _{1}x)²(y - m_{2}x)² | ||

= EG(x, y)FH(x, y)F'G'(x, y)E'H'(x, y) | ||

= 0. |

describes the pencil of quartic curves that pass through the eight points E, F, G, H, E', F', G' and H'. Further, there exist some constant λ and μ such that a degenerate quartic curve of four straight lines passing through these points is uniquely determined. Hence we can choose the constants λ and μ such that the degenerate quartic function

f(x,y) | = λ[b(x + a) + (e - a)y][bd(x - c) + (de - c)y][b(x - c) + (c + e)y][bd(x + a) + (de + a)y] + μ(y - m _{1}x)²(y - m_{2}x)² | |

= EG(x, y)FH(x, y)F'G'(x, y)E'H'(x, y) | ||

= 0. |

Now it can be seen in the figure that the line y = 0 cuts the four lines EG, FH, F'G' and E'H' at the points M, N, M' and N' respectively.

Hence the equation

f(x, 0) = λb^{4}d^{2}(x + a)^{2}(x - c)^{2} + μm_{1}^{2}m_{1}^{2}x^{4} = 0

has four real roots {IM, IN, -IM', IN'} if λ and μ are oppositely signed. Setting _{1}^{2}m_{1}^{2} /λb^{4}d^{2} = -k^{2}

(x + a)^{2}(x - c)^{2} - k^{2}x^{4} = 0.

Factoring the left-hand side gives two quadratic equations:

(1 + k)x² + (a - c)x - ac = 0 and

(1 - k)x² + (a - c)x - ac = 0.

Now it is obvious that the roots, say, x and x' of either of the two quadratic equations satisfy the following relation

1/x + 1/x' = (x + x')/xx' = -(a - c)/(1 ± k)/(-ac)/(1 ± k) = 1/c - 1/a.

Therefore it can be concluded that

(1) | 1/IM - 1/IA = 1/IN - 1/IC. |

Now, by symmetry, it can be easily seen that in the figure, the intersections P and Q of lines EH and FG with the diagonal BD respectively have a similar property such that

(2) | 1/IP - 1/IQ = 1/IB - 1/ID = constant. |

The quadrilateral ABCD itself can be seen as composed of two butterflies ACDBA and BDACB which both with their wing tips touching both diagonals AC and BD satisfy both equations (1) and (2). If we connect intersections inside the quadrilateral and extend lines to create intersections outside it we can get infinitely many butterflies which all have the properties of being invariant under the equations (1) and (2).

From the proof above it can be seen that the quadrilateral doesn’t have to be convex. Actually, butterflies inhabit in any quadrilateral defined by four points in general position, only the distances in equations (1) and (2) are signed according to the orientation of the line.

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