The Lepidoptera of the Triangle

Sidney Kung
July, 2107

Theorem 1

The Lepidoptera of the Triangle, Theorem 1, whole

Proof of Theorem 1

We begin by assuming a more general case than that shown in Figure 1, in that

  1. none of the three transversals $DE\,$ $FG,\,$ and $HJ\,$ needs to be parallel to a side of $\Delta ABC,$
  2. we only ask that the points $P=AB\cap FG,\,$ $Q=HJ\cap BC,\,$ $R=DE\cap AC\,$ be collinear.

The Lepidoptera of the Triangle, proof of Theorem 1

Denote the six vertices of the hexagon $HDFJEG\,$ in the order $325416.\,$ We see that the points of intersections of the three pairs of the opposite sides $23-56,\,$ $34-61,\,$ $12-45\,$ are, respectively $P,\,$ $Q,\,$ $R.\,$ Thus, by the converse of Pascal's Theorem, if $P,\,$ $Q,\,$ $R\,$ are collinear, the hexagon $HDFJEG\,$ is inscribed in a conic.

Return to Figure 1. Denote the vertices of the hexagon $HDFJEG\,$ in the order $325416.\,$ Since $FG\parallel AB,\,$ $23-56\,$ meet at a point of infinity. Similarly, the same holds for $34-61\,$ and $12-45.\,$ Thus, the points $P,\,$ $Q,\,$ $R,\,$ all being points at infinity, are collinear, hence the hexagon $HDFJEG\,$ is inscribed in a conic. It follows then from Candy's generalized Butterfly Theorem, that

$\displaystyle \frac{1}{HI}-\frac{1}{IJ}=\frac{1}{IX}-\frac{1}{IY},$

which is (1), and (2) is immediate if $I\,$ is the midpoint of $HJ.$

Theorem 2

With a reference to Figure 3i, three lines $DE,\,$ $FG,\,$ $HJ\,$ are drawn through point $I\\,$ in $\Delta ABC,\,$ tangent to the circumcircles $(AIC),\,$ $(BIA)\,$ $(BIC),\,$ respectively.

The Lepidoptera of the Triangle, Theorem 2 and proof

Let $P=AB\cap FG,\,$ $Q=HJ\cap BC,\,$ $R=DE\cap AC.\,$ If $DG\,$ and $EF\,$ intersect $HJ\,$ at $X,Y,\,$ respectively, then


$\displaystyle \frac{1}{HI}-\frac{1}{IJ}=\frac{1}{IX}-\frac{1}{IY}.$


If $HI=IJ\,$ then $IX=IY.$

Proof of Theorem 2

Consider $DR\,$ and the circumscribe circle of $\Delta AIC$ (see Figure 3ii). It's easy to see that $\Delta RIA\sim\Delta RCI.\,$ So,

$\displaystyle \frac{RA}{RI}=\frac{IA}{IC},\;\frac{RC}{RI}=\frac{AC}{IA}.$

Dividing the former by the latter, we get $\displaystyle \frac{RA}{RI}=\frac{IA^2}{IC^2}.\,$ Similarly, since $\Delta HQC\sim\Delta QBI\,$ and $\Delta PIC\sim\Delta PBI,\,$ $\displaystyle \frac{QC}{QB}=\frac{IC^2}{IB^2}\,$ and $\displaystyle \frac{PB}{PA}=\frac{IB^2}{IA^2}.\,$ Thus

$\displaystyle \frac{PB}{PA}\cdot\frac{QC}{QB}\cdot\frac{RA}{RC}=\frac{IB^2}{IA^2}\cdot\frac{IC^2}{IB^2}\cdot\frac{IA^2}{IC^2}=1.$

By the converse of Menelaus' theorem, the three points $P,Q,R\,$ are collinear, implying that the hexagon $HDFJEG\,$ is inscribed in a conic. Consequently, Candy's theorem delivers (3) and (4).

Butterfly Theorem and Variants

  1. Butterfly theorem
  2. 2N-Wing Butterfly Theorem
  3. Better Butterfly Theorem
  4. Butterflies in Ellipse
  5. Butterflies in Hyperbola
  6. Butterflies in Quadrilaterals and Elsewhere
  7. Pinning Butterfly on Radical Axes
  8. Shearing Butterflies in Quadrilaterals
  9. The Plain Butterfly Theorem
  10. Two Butterflies Theorem
  11. Two Butterflies Theorem II
  12. Two Butterflies Theorem III
  13. Algebraic proof of the theorem of butterflies in quadrilaterals
  14. William Wallace's Proof of the Butterfly Theorem
  15. Butterfly theorem, a Projective Proof
  16. Areal Butterflies
  17. Butterflies in Similar Co-axial Conics
  18. Butterfly Trigonometry
  19. Butterfly in Kite
  20. Butterfly with Menelaus
  21. William Wallace's 1803 Statement of the Butterfly Theorem
  22. Butterfly in Inscriptible Quadrilateral
  23. Camouflaged Butterfly
  24. General Butterfly in Pictures
  25. Butterfly via Ceva
  26. Butterfly via the Scale Factor of the Wings
  27. Butterfly by Midline
  28. Stathis Koutras' Butterfly
  29. The Lepidoptera of the Circles
  30. The Lepidoptera of the Quadrilateral
  31. The Lepidoptera of the Quadrilateral II
  32. The Lepidoptera of the Triangle
  33. Two Butterflies Theorem as a Porism of Cyclic Quadrilaterals
  34. Two Butterfly Theorems by Sidney Kung

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