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

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