The Lepidoptera of the Quadrilateral II

Sidney Kung
April 26, 2008

We prove the Butterfly Theorem with the aid of Menelaus' theorem.

Through the intersection I of the diagonals AC, BD of a convex 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.

Proof

 
 Figure 1.

Refer to Figure 1. Triangles BAD and CAD are cut by transversal EIF. So, by applying Menelaus' theorem twice, we have

BE/EA · AK/KD · DI/IB = 1,
CF/FD · DK/KA · AI/IC = 1.

Multiplying the above gives

BE/EA · ID/IB · CF/FD · IA/IC = 1,

or

(1) IA · ID · BE/EA = IC · IB · FD/CF.

Similarly, since transversal HIG cuts triangles BDC and ADC, applying Menelaus' theorem twice again will give us

AG/GD · DI/IB · BH/HC · CI/IA = 1,

or

(2) ID · IC · BH/HC = IB · IA · DG/GA.

Note that neither (1) nor (2) involves IM, MA, IN, or NC. The relationship among these and other line segments in the figure will be established as follows:
 
 Figure 2.

Refer to Figure 2. Let L and P be the intersections of FH and EG with BG. Consider the triangle CDI with transversal FH, we have

DF/ FC · CN/NI · IL/LD =1,

or

(3) DF/FC = IN/NC · LD/IL.

Similarly, from triangle CIB, we see that

BH/HC · CN/NI · IL/LB = 1,

or

(4) BH/HC = NI/CN · LB/IL.

Thus,

(5)
IB · DF/ FC + DI · BH/HC= IB · IN/NC · LD/IL + DI · NI/CN · LB / IL
 = NI/CN · (IB · ID + DI · LB) / IL
 = NI/CN · (IB · (DI + IL) + DI · (IL- IB)) / IL
 = DB · IN/NC.

Again, we apply the Menelaus theorem on triangles AIB and ADI to get

(6) BE/EA = MI / AM · PB/IP

and

(7) DG / GA = MI / AM · PD/IP,

respectively. So,

(8)
IB · DG/GA + ID · BE / EA= MI/MA (IB · PD + PB · ID) / IP
 = MI/MA · (BI · (IP - ID) + ID · (PI + IB)) / IP
 = BD · IM/MA.

Now, multiplying (5) by IC, we get

(9) IC · IB · DF/FC + IC · ID · BH/HC = IC · BD · IN/NC

and multiply (8) by IA, we have

(10) IA · IB · DG/GA + IA · ID · BE/EA = IA · BD · IM/MA.

Finally, by subtracting (10) from (9), and taking into account (1) and (2), we obtain

BD · ((IC · IN)/NC - (IA · IM)/MA ) = 0.

It follows then that

(IA - IM)/(IA · IM) = (IC - IN) / (IC · IN).

Therefore, 1/IM - 1/IA = 1/IN - 1/IC, as required

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
  35. Butterfly in Complex Numbers

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