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) 

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) 

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
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