Research on Butterfly Theorem Butterfly Theorem is one of the most appealing problems in the classic Euclidean plane geometry. The name of Butterfly Theorem is named very straightforward that the figure of Theorem just likes a butterfly. Over the last two hundreds, there are lots of research achievements about Butterfly Theorem that arouses many different mathematicians’ interests. Until now, there are more than sixty proofs of the Butterfly Theorem, including the synthetical proof, area proof, trigonometric proof, analytic proof and so on. And based on the extension and evolution of the Butterfly Theorem, people can get various interesting and beautiful results. The definition of the Butterfly Theorem is here below: “Let M be the midpoint of a chord PQ of a circle, through which two other chords AB and CD are drawn; AD cuts PQ at X and BC cuts PQ at Y. Prove that M is also the midpoint of XY.” (Bogomolny) This is the most accurate definition currently. However, Butterfly Theorem has experienced some changes and developments. The first statement of the Butterfly Theorem appeared in the early 17th century. In 1803, a Scottish mathematician, William Wallace, posed the problem of the Butterfly Theorem in the magazine The Gentlemen’s Mathematical Companion. Here is the original problem below: “If from any two points B, E, in the circumference of a circle given in magnitude and position two right lines BCA, EDA, be drawn cutting the circle in C and D, and meeting in A; and from the point of intersection A to the centre of the circle AO be drawn, and the points E, C; B, D joined, and produced to meet an indefinite perpendicular erected at A on AO; then will FA be always equal AF. Required the demonstration?”(Bogomolny) (Figures of W Wallace’s question) Soon afterwards, there were three solutions published in 1804. And in 1805, William Herschel, a British astronomer, wrote a letter to Sir William Wallace to raise the question again. Here is the content of the letter: “I have kept a little problem for you which a friend of mine has sent me who says he cannot find a solution of it. I mentioned to him that I had a friend who would probably help him to one. The problem is this. Given AB the diameter of a circle. CD a chord cutting it at right angles in K. EF, and HG two other chords drawn any how through the point K; and HF, EG chords joining the extremes of EF, HG. Required to prove that MK is equal to LK.” (Bogomolny) (Figure of William Herschel’s question) In 1814, Rev. Thomas Scurr also posed the same question in the magazine Gentlemen’s Diary or Mathematical Repository. Here is his question: “Let AB be any chord in a circle, through I the middle point of which let two other chords be drawn, the former MIN, meeting the circle in M, N, the latter OIL, meeting the circle in O, L; join the points L, M, and N, O, by the right lines which cut the chord AB in P and Q respectively; then is AP=BQ. Required a demonstration.” (Bogomolny)
(Figure of Rev. Thomas Scurr’s question) After the question posed on Gentlemen’s Dairy or Mathematical Repository, there were many different and excellent proofs provided by various mathematicians. In the same year of question posed, William George Horner who was known as Horner’s method provided the first proof, and Richard Taylor also provided the different proof. In 1827, Miles Bland gave his proof in his book Geometrical Problems. Other than the proofs by basic and traditional methods, there were also some mathematicians came up with the new and interesting methods of geometry. In 1904, John Casey provided the most succinct proof of Butterfly Theorem in his book A Sequel to the First Six Books of the Elements of Euclid. He used projective geometry, a branch of classic geometry, and he mentioned that he just used the cross ratio of pencils of lines to prove. And in 1981, the magazine Crux published Kesirajn Satyanarayana’s proof of Butterfly Theorem. He used the theorems of the family of lines and pencil of conics in the area of analytical geometry, which was also a simple method relatively. During the two hundreds years, Butterfly Theorem motivated many mathematicians or enthusiasts to study the various methods of geometry or even other mathematical field such as algebra to solve the Butterfly Theorem. With the development of Butterfly Theorem, people also find that Butterfly Theorem can be applied on different geometric figures including ellipse, hyperbola, parabola, quadrilateral, and many others. Also, there are many derived theorems and corollaries were inferred in order to prove the Butterfly Theorem on other geometric figures. Therefore, Butterfly Theorem flourishes the study of geometry and helps develop the mathematics. It is very worth to explore some significant and interesting proofs of Butterfly Theorem. In this paper, I would like to present proofs of William Wallace, William George Horner, and Butterfly Theorem in Hyperbola and different quadrilaterals. Firstly, it is very important to study the solution given by William Wallace to William Herschel’s question because William Wallace is the first person to give the proof of Butterfly Theorem. And recently, the original solution of William Wallace was posed in the magazine Journal of the British Society. And here is the process of his proof: 1. Draw a parallel PQ such that passes the point L and intersects line KF at P and KH at L. (I31) 2. HQP=HGE. (I29) 3. HGE=HFE or HFP. (Book3, I21 that in a circle the angles in the same segment equal on another) 4. HQP=HFP. (CM1) 5. GEK=QPK. (I29) 6. GEK=FHK. (Book3, I21) 7. QPK=FHK. (CM1) 8. QKP=FKH. 9. QKP~FKH. (Def. of similar triangles) 10. PL*LQ=HL*LH. 11. EKG=PKQ. (Def. of vertical angles) 12. EGK=PQK. (I29) 13. KEG~KPQ. (5,11,12, Def. of similar triangles) 14. KM^2=EM*MG=KL^2=PL*QL or FL*LH 15. KM^2+CM*MD=KL^2+CL*LD 16. KM^2+CM*MD=CK^2 17. KL^2+CL*LD=KD^2 18. KM^2/CK^2=KL^2/KD^2 (16, 17, algebra) 19. KM/CK=KL/KD (18, algebra) 20. CD is perpendicular to diameter (hypothesis) 21. KC=KD. 22. KM=KL. And William Wallace also gave a remark:
“The above proposition is a particular case of a more general one extending to all the Conic Sections, which may be expressed thus. If AB is any diameter of a Conic Section, and CD any right line cutting it in K, and parallel to a tangent at its vertices; also EF and HG two other lines drawn anyhow through K, to meet the conic section, the one in the points E, F, and the other in the points G, H, the straight lines EG, FH which join the extremities of these lines shall intercept upon CD the segments KM, KL which are equal to each other. ” (Crack&Connor)
(Figure of W Wallace’s proof)
In William Wallace’s proof, it is easy to find that he mainly used the definition and theorems of similar triangles. And he also found that Butterfly Theorem could be applied to all the conic sections, instead of just circle. Looking back to his question posed on The Gentlemen’s Mathematical Companion, the question is more general than William Herschel’s and Rev. Thomas Scurr’s. In William Wallace’s question there are totally three posiblities. Firstly, there are mainly two cases for the perpendicular from A to AO that the perpendicular crosses or does not cross the circle O. If the perpendicular crosses the circle O, there are also two possibilities for the intersections G and F that fall within or outside the circle O. and for the perpendicular does not cross the circle, there is just one case that intersections G and F fall outside the circle. Therefore, William Wallace’s idea helps other mathematicians to have open minds on study of Butterfly Theorem.
In 1814, William George Horner, who was recorded as the first person proved the Butterfly Theorem before William Wallace’s proof was found, firstly submitted his solution of Rev. Thomas Scurr’s question posed on Gentlemen’s Diary or Mathematical Repository. And here is Horner’s proof according to Coxeter and Greitzer’s record: 1. Drop perpendiculars x1, x2 from point X to segment AM, DM with foots E, F respectively and drop perpendiculars y1, y2 from point Y to segment BM and CM with foots G, H respectively. (Theorem 4.1.3) 2. XEM=MGY=MHY=XFM=90. (Def. of perpendicular) 3. EMX=GMY and HMY=FMX. (Theorem 3.5.13) 4. MXE=MYG and MXF=MYH. (Angle Sum Theorem) 5. Then MEX~MGY and MFX~MHY. (2, 3, 4, def. of similar triangles) 6. x/y=x1/y1 and x/y=x2/y2. (Theorem 5.3.1) 7. XEA=YHC=XFD=YGB=90. (2, linear pair) 8. XAM=YCM and CDM=YBM. (Book3, I21) 9. Then AXE=CYH and DXF=BYG. (Angle Sum Theorem) 10. Then XAE~YCH and XDF~YBG. (7, 8, 9, def. of similar triangles) 11. x1/y2=AX/CY and x2/y1=XD/YB. (Theorem 5.3.1) 12. x^2/y^2=(x1/y2)*(x2/y1)=(x1/y1)*(x2/y2)=(AX/XD)*(CY*YB)=(PX*XQ)/(PY*Y1). (6, 11, algebra) 13. Then x^2/y^2=((a-x)(a+x))/((a-y)*(a+y))=(a^2-x^2)/(a^2-y^2)=a^2/a^2=1. (12, algebra) 14. Then x=y. (13, algebra) (Figure of Horner’s proof) From Horner’s proof, we can find he also used the property of similar triangle and the basic algebra as well as William Wallace, while his proof is more succinct because he also used the property of perpendicular to make the proof easier. Another reason I think is that William Herschel’s question is more general than Rev. Thomas Scurr’s. Rev. Thomas Scurr emphasized the midpoint of a chord in his problem, while William Herschel just made a chord to be the diameter. And current Butterfly Theorem corresponds more to Rev. Thomas Scurr’s problem. With the continuous study of Butterfly Theorem and open mind of the human, there are lots of extensions of application of Butterfly Theorem on other geometric figures, instead of just the circle. The professor of University of North Florida, Sidney H. Kung, has studied and proved the Butterfly Theorem in hyperbola and quadrilaterals. In order to apply the Butterfly Theorem in the hyperbola, the theorem should be corresponding to the property of hyperbola: “Let (0,k) be the midpoint of a chord AB parallel to the major axis of a hyperbola. Through M two other chords CD and EF are drawn. ED cuts AB at P and CF cuts AB at Q. Then M is also the midpoint of PQ.” (Bogomolny) And here is Sydney H. Kung’s proof below: 1. Set up an x-y-axes such that M is the original point (0,0) 2. And we have the equation of hyperbola that x2a2-y+k2b2=1 or b2x2-a2y+k2-a2b2=0. 3. We have points M(0,0), C(x1,y1), D(x2,y2), E(x3,y3), F(x4,y4), P(p,0), Q(q,0) 4. Then we have linear equations y=m1x and y=m2x of lines DC and EF respectively. 5. Substituting y=m1x into hyperbola equation a2y+k2-a2b2=0.
Then we get a new equation b2-a2m12x2-2a2m1kx-a2b2+k2=0. 6. We have x1+x2=--2m1ka2b2-m12a2 for roots x1 and x2.
Then dividing the equation of step 5 by (x1+x2), we get m1x1x2x1+x2=-(b2+k2)2k. 7. Using the same method by substituting y=m2x into a2y+k2-a2b2=0. 8. Then we get m2x3x4x3+x4=-(b2+k2)2k. 9. Then m1x1x2x1+x2=m2x3x4x3+x4. 10. Then –x2x3m1x2-m2x3=x1x4m1x1-m2x4. 11. Since points C, Q, and F are collinear. 12. Then slopes of QC and FQ are equal. 13. Then y1x1-q=-y4q-x4.
Then q-x1q-x4=y1y4=m1x1m2x4 such that q≠x1 and q≠x4. 14. Then we have q=(m1-m2)x1x4m1x1-m2x4. 15. Same steps with another case that slopes of PE and DP are equal. 16. Then we get p=(m1-m2)x2x3m1x2-m2x3 17. Then we have p=q. (10, 14, 16) 18. Then we have MP=MQ.
(Figure of Butterfly Theorem in hyperbola)
Kung’s proof of Butterfly Theorem in hyperbola is more complicated than other’s proof in a circle and he uses the algebraical method to prove. Since the proof in hyperbola is more difficult than it in circle, using the property of cartesian coordinates will help the proof easier.
Another interesting proof by Sydney H. Kung is proving the Butterfly Theorem in quadrilaterals. It is very necessary to mention that Butterfly Theorem is just able to be applied in the convex quadrilateral. Before the proof of Butterfly Theorem in quadrilateral, it is important to know two properties of areas of triangles that will be used in proof because Sydney H. Kung uses the area method to prove. Two properties are following:
“P1: If K is the intersection of the lines XY and UV, VK, then A(UXY)A(VXY)=UKVK , where A(UXY) denotes the area of triangle UXY.”
“P2: Given triangles ABC and XYZ, suppose that, we have ABC=XYZ or ABC+XYZ=180, then A(ABC)A(XYZ)=AB*BCXY*YZ.” (Kung) (Figure of P1)
(Figure of P2)
Then we need to change the Butterfly Theorem to correspond to the quadrilaterals:
“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 at E, F, G, H. If M=EGAC, N=HFAC, then AMIM*INCN=IAIC .” (Kung)
And here is the proof following: 1. Let ☐ABCD be a convex quadrilaterals, AC and BD be diagonals such that intersects at I, EF be a line through I intersects with AB and CD at E and F respectively, and GH be a line through I intersects with AD and BC at G and H respectively. (Hypothesis) 2. Draw the lines AH, AF, CE, CG, EF, and GH. (Incidence postulate) 3. Then we get twelve pairs of triangles AEG and IEG, IFH and CHF, IFH and IEG, CBD and CHF, ABD and CBD, AEG and ABD, AFC and EAC, HAC and GAC, DAC and FAC, BAC and HAC, EAC and BAC, GAC and DAC, and each pair of these has a common side or a common angle. 4. By P1 and P2, we can get that AMIM*INCN=AAEGAIEG*AIFHACHF=AIFHAIEG*ACBDACHF*AABDACBD*AAEGAABD=IE*IHIE*IG*CD*CBCF*CH*IAIC*AE*AGAB*AD=AAFCAEAC*AHACAGAC*ADACAFAC*ABACAHAC*IAIC*AEACABAC*AGACADAC =IAIC. 5. Let IA=a, IC=c, IM=m, IN=n. 6. Then a-mm*nc-n=ac. 7. Then 1m-1n=1a-1c . 8. Since the properties of Butterfly Theorem in the circle are same as that in the convex quadrilateral. 9. Then the ratios (AMIM)(CNIN) and IAIC are equal. 10. Then harmonic mean of IC and IM= harmonic mean of IA and IN. 11. Then we have IC=IA or IM=IN.
(Figure of Butterfly Theorem in quadrilateral)
I have mentioned that the Butterfly Theorem just can be applied in the convex quadrilateral. Through the proof of Sydney Kung, we can find that the quadrilateral contained the butterfly must have the same relation as the circle. Therefore, the quadrilateral should be cyclic that the vertices of quadrilateral all lie on a circle. Since one of the four vertices of the nonconvex circle is the interior of the angle of other three vertices, the nonconvex quadrilateral could not be cyclic. After proving the Butterfly Theorem for convex quadrilaterals, we could also easily prove the Butterfly Theorem in parallelograms, trapezoids, rhombus and other convex quadrilaterals that have the special properties. As a classic and historical proposition, the Butterfly Theorem has already composed a big “Butterfly family” in plane geometry. There are many derived extensions and generalizations that make many mathematicians and enthusiastics very excited. Various proofs using by different methods make the mathematical field develop dramatically. During the study of the Butterfly Theorem, there are also many lemmas produced that can be applied to other geometric problems, instead of just the Butterfly Theorem. Other than the proofs I presented above, there are still many beautiful proofs that are worth to study. We can also imagine that the extension of the Butterfly Theorem could be wider and even be applied in the space geometry.

References
A. Bogomolny, “William Wallace's 1803 Statement of the Butterfly Theorem.” Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/pythagoras/WallaceButterfly.shtml, Accessed 24 May 2015
A. Bogomolny, “William Wallace's Proof of the Butterfly Theorem.” Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/pythagoras/WilliamWallaceButterfly.shtml, Accessed 24 May 2015
Alex D D Craik and John J O’ Connor, “Some unknown documents associated with William Wallace (1768-1843).”, Journal of the British Society for the History of Mathematics, 26:1, 17-28.
A. Bogomolny, “Butterfly theorem.” Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#CG, Accessed 25 May 2015
A. Bogomolny, “Butterflies in Hyperbola.” Interactive Mathematics Miscellany and Puzzles http://www.cut-the-knot.org/pythagoras/ButterflyInHyperbola.shtml, Accessed 25 May 2015
Sidney Kung. “A Butterfly Theorem for Quadrilaterals.” Mathematics Magazine 78.4 (2005): 314-6. ProQuest. Web. 26 May 2015.