butterfly theorem 1

8/12/2019 Butterfly Theorem 1

1/22

The Butterfly Theorem

According to Coxeter and Greitzer, one of the solutions to the Butterfly theorem was

submitted in 1815 by W. G. Horner of Horner's methodfame. Most recently a 1805 proof by

William Wallace has been discoveredin Wallace's family archives.

The solutions I gathered below may serve as a basis for a discussion of which proofs better

clarify the gist of the problem, if at all. Why is the result true? (See the discussions on

Concyclic Circumcenters: Dynamic ViewandA Sequel.)

Theorem

Let M be the midpoint ofa chord PQof 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.

Proof 0 (William Wallace, 1805)

William Wallace's proof is based on the following diagram:
http://www.cut-the-knot.org/ctk/SixCircum.shtmlhttp://www.cut-the-knot.org/ctk/SixCircum2.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#CGhttp://www.cut-the-knot.org/ctk/SixCircum2.shtmlhttp://www.cut-the-knot.org/ctk/SixCircum.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#COChttp://www.cut-the-knot.org/Curriculum/Calculus/HornerMethod.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#CG


2/22

I have placed it on a separate page.

Proof 1 [Coxeter and Greitzer, p. 45]

Let's drop perpendiculars x1and x2from X and y1and y2fromY on AB and CD. Let's also

introduce a = MP = MQ and x = XM and y = YM. Observe

several pairs of similar triangles, which implies the following

proportions:

Triangles Proportion

Mx1and My1 x/y = x1/y1

Mx2and My2 x/y = x2/y2

Ax1and Cy2 x1/y2= AX/CY

Dx2and By1 x2/y1= XD/YB

from which
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#CGhttp://www.cut-the-knot.org/pythagoras/WilliamWallaceButterfly.shtml


3/22

x/y = x1/y2 x2/y1= x1/y1 x2/y2= AXXD/ CYYB = PXXQ/ PYYQ.

So that

x/y = (a - x)(a + x)/ (a - y)(a + y) = (a - x)/(a - y) = a/a = 1.

And finally x = y.

Proof 2 [Shklyarsky, problem 104, solution 1]

Let O be the center of the given circle. Since OM XY, in

order to show that XM = MY, we have to prove

that XOM = YOM.

Drop perpendiculars OK and ON from O onto AD and BC,

respectively. Obviously, K is the midpoint of AD and N is the

midpoint of BC. Further,

DAB = DCB and ADC = ABC,

as angles subtending equal arcs. Triangles ADM and CBM are therefore similar,

and AD/AM = BC/CM, orAK/AM = CN/CM. In other words, in triangles AKM and CNM two pairs

of sides are proportional. Also, the angles between the corresponding sides are equal. We

infer that the triangles AKM and CNM are similar. Hence, AKM = CNM.

Now, have a look at the quadrilaterals OKXM and ONYM. Both have a pair of opposite straight

angles, which implies that both are inscribable in a circle. In OKXM, AKM = XOM. In

ONYM,CNM = YOM. From which we get what we've been looking for: XOM = YOM.

Proof 3 [Shklyarsky, problem 104, solution 2]

For convenience, denote the angles as on the diagram on the right. Let x = XM and a = PM. As

in Proof 1,

AXXD = PXXQ
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#SHhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#SH


4/22

= a - x.

In DXM, by the Law of Sines, we have

DX; = xsin()/sin(180 - ( + + ))

= xsin()/sin( + + ).

And in AXM

AX = xsin()/sin(),

which leads to

AXDX = xsin()sin()/sin()sin( + + ) = a - x.

From here we may find x:

x = asin()sin( + + ))/(sin()sin() + sin()sin( + + )).

The latter expression is symmetric in and . Therefore, if we repeat the derivation for the

segment y = MY, we'll get exactly same result. Hence, x = y.

Proof 4

This proof originates in Projective geometry, but could be explained without invoking the

theory. (It's loosely based on problem 29.46 from [Problems].) Let S be the given circle with

center O. Assume for a moment that there exists a (projective) transformation T of the plane

that

1. maps S onto another circle S',

2. moves M onto the center O' of S',

3. preserves the ratio of segments in the direction perpendicular to OM,

4. transforms AB, CD, AD, BC, and PQ onto the segments A'B', C'D', A'D', B'C', and P'Q' with

5. P'Q' still perpendicular to OM.

(I'll prove the existence of T shortly.) A'B', C'D' and P'Q' serve as diameters of S'. O' is the
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#VP3http://www.cut-the-knot.org/proofs/sine_cosine.shtml#law


5/22

center of a half-turn that takes A' into B', C' into D', P' into Q', and, therefore, X' into Y',

where X' and Y' are images of X and Y. Triangles A'X'O' and B'Y'O' are then equal, so

that X'O' = O'Y'. But, by #3 and #5 above, XM/MY = X'O'/O'Y'. Therefore, too, XM = MY.

It remains to be shown that the transformation T with properties 1-5 does exist. I'll construct

T as a composition of two transformations: a central projection and an affine mapping.

Let UV be the diameter of S through M. UV PQ. Imagine looking at the circle from a side on

the level of its plane in the direction of PQ. We see a segment UV with the midpoint at O and

another point M between O and U. Choose a point K on a perpendicular to UV at O. Draw KM.

Find a line VNW such that VN = NW. (This is simple: just draw a line through O parallel to KU.

Let N be the point of intersection of that line with AM. Extend VN to the intersection with AU

at W.)

Now, think of K as the apex of a right cone with base S. VW is then an axis of the ellipse cut

from the cone by a plane through VW perpendicular to the plane of the drawing. N is its

center. The central projection I mentioned above maps M to N, U to W, V to V, and, in

general, a point Z in the plane of S to the point of intersection of KZ with the plane of the

ellipse. The latter could now be squeezed in the direction of VW so as to map the ellipse

onto a circle keeping N as its center.

It's quite obvious that the composition of the two transformations satisfies 1-5 above.
http://www.cut-the-knot.org/triangle/pythpar/Geometries.shtml#affine


6/22

Proof 5 [Essays, p. 59]

The proof is very simple but requires familiarity with a general notion of conicsand second

degree equations that describe the conics in Cartesian coordinates.

In the plane Cartesian coordinates conics are described by second degree equations:

f(x, y) = Ax + Bxy + Cy + Dx + Ey + F = 0.

Any (non zero) multipleof such an equation, describes exactly the same conic, which is to

say that it takes 5 numbers to describe a conic, not 6 as might appear from the equation. A

conic is thus uniquely determined by 5 distinct points.

Straight linesare described by linear equations

l(x, y) = Ax + By + C = 0.

Straight lines are uniquely determined by two points. Below, for a point X with

coordinates (x, y),l(x, y) and l(X) are used interchangeably, depending on which is more

convenient under the circumstances. With these preliminaries we may prove the following

Lemma

Let the points P, Q, R, and S lie on a conic given by a second degree equation f = 0. Then, for

some constant and ,

f = lPQlRS+ lPSlQR,

where lPQ= 0, lRS= 0, lPS= 0, and lQR= 0 are the (linear) equations of PQ, RS, PS and QR,

respectively.

Proof of Lemma

Let U be a fifth point on the conic at hand. Then surely there exist 0and 0such that

0lPQ(U)lRS(U) + 0lPS(U)lQR(U) = 0.

Define a second degree expression
http://www.cut-the-knot.org/Curriculum/Calculus/StraightLine.shtmlhttp://www.cut-the-knot.org/do_you_know/mul_eq.shtmlhttp://www.cut-the-knot.org/do_you_know/few_words.shtml#Cartesianhttp://www.cut-the-knot.org/proofs/conics.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#VP1


7/22

f0(X) = 0lPQ(X)lRS(X) + 0lPS(X)lQR(X).

Clearly f0(U) = 0. But also f0(P) = f0(Q) = f0(R) = f0(S) = 0. (For

example, f0(P) = 0 because lPQ(P) = 0and lPS(P) = 0.) The equation f0(X) = 0 defines the same

conic as f = 0, so that f and f0are multiples of each other: f = f0for some . Which proves

the lemma.

Let's now continue the proof of the Butterfly theorem. Let f = 0 be an equation of the given

circle. Then by lemma, for some and

f = lABlCD+ lADlBC.

This equation also holds for the restrictions of all the functions involved onto the line PQ.

Take PQ as the x-axis with the origin at M. Then, since f(P) = f(Q) = 0 and PM = MQ, we can

assume thatf(x, 0) = x - a and lABlCD= bx on PQ. If so, restricted to PQ, lADlBC= cx - d. This

means that the two roots of the equation lADlBC= 0 are equidistant from the origin M.

Proof 5'

The Butterfly theorem was offered as problem A6 at the 24th Putnam competition (1963). The

solution I came across on the Web, is a modification of Proof 5based on a slightly more

general lemma: Let g = 0 and h = 0 be equations of two distinct conics through 4 distinct

points. Then any conic through the four points has the equation g + h = 0, for some and .

The circle is then defined, by, say, g = 0 and the pair of lines AB, CD by h = 0. Then the pair

of lines AD, BC is another conic through the same four points and satisfies f = g + h = 0, for

some and . Now, g and h have symmetric (with respect to M) restrictions to PQ, and so has

f.

Remark
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Proof5


8/22

Note that the latter proof only requires that the restrictions of AB and CD on PQ be symmetric

with respect to M. This means that the equality PX = QY holds not only when AB and CD pass

through M, but also when they cross PQ at points equidistant from M! We thus get a simple

proof of the generalization of the Butterfly theorem first established by M. Klamkin in 1965

(Mathematics Magazine, Volume 38, Issue 4 (Sep., 1965), 206-208.)

Proof 6

The Butterfly theorem is a particular case of a more general Two Butterfliestheorem.Choose one "butterfly quadrilateral" to be symmetric with respect to OM and so

that its intersecting sides meet at M. The second "butterfly quadrilateral" is bound to intersect

PQ at the same points as the first, i.e., symmetrically with respect to M.

Proof 7

Another generalizationof the theorem was pointed out to me by Qiu FaWen, a Chinese

teacher who discovered the result with his students in 1997. The Chinese version is available

athttp://www.qiusir.com/report2000/.htm. I had to do some guess

work to figure out what it was about.

The Butterfly theorem follows trivially from the generalization when the two circles involved

coalesce into one.
http://www.qiusir.com/report2000/%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD.htmhttp://www.cut-the-knot.org/pythagoras/BetterButterfly.shtmlhttp://www.cut-the-knot.org/Curriculum/Geometry/TwoButterflies.shtml


9/22

Proof 8

This proof is by Steve Conrad as appeared in Samuel Greitzer's Arbelos.

The basic point of departure is the observation that, if in two triangles an angle of one is

equal to an angle of the other, then the areas of the triangles are in the same ratio as the

products of the sides composing the equal angles. In the diagram above, there are four pairs

of equal angles, which allows us to write four equalities:

1: Area(XAM)/Area(MCY) = AXAM / CMCY

2: Area(CMY)/Area(DMX) = CMMY / DMMX

3: Area(XDM)/Area(MBY) = DXDM / BMBY

4: Area(BMY)/Area(AMX) = BMMY / AMMX

Now, since the product of the ratios on the left equals 1, the same is true for the product of

the right hand sides, which after some cancellation appears as

AXDXMY / CYBYMX = 1,

or

(1) AXDX / CYBY = MX/ MY.

By the power of a point theorem,

AXDX = PXQX = (MP - MX)(MQ + MX)
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Arbelos


10/22

and similarly

CYBY = QYPY = (MQ - MY)(MP + MY).

Hence, from (1)

(2) (MP - MX)(MQ + MX) / MX = (MQ - MY)(MP + MY)/ MY.

Since MP = MQ, we obtain

MP/ MX - 1 = MP/ MY - 1,

which implies MX = MY.

Remark

Without the assumption MP = MQ, (2) simplifies to

MPMQ(MX - MY) = MXMY(MP - MQ).

The latter is the essence of A. Candy's theorem (1896), which is usually written as

1/MP - 1/MQ = 1/MX - 1/MY.

As it turns out, this is very close to Candy's original derivation.

Proof 9

Following is a proof of Klamkin's generalizationwith the aid ofMenelaus' theorem.
http://www.cut-the-knot.org/Generalization/Menelaus.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#gb


11/22

Triangle XSY is cut by two transversals, CZD and AWB such that we may apply Menelaus'

theorem twice:

(AWB) XA/SA SB/YB YW/XW = 1

(CZD) XD/SD SC/YC YZ/XZ = 1

The product of the two is

(3) XAXD/SASD SBSC/YBYC YZ/XZ YW/XW = 1.

From the Power of the Point theorem, SASD = SBSC, so that (3) simplifies to

(4) XAXD/YBYC YZ/XZ YW/XW = 1.

By the same theorem XAXD = PXQX and YBYC = PYQY. Therefore (4) becomes

(5) PXQX/PYQY YZ/XZ YW/XW = 1.

Let PM = MQ = a, ZM = MW = b, XM = x, YM = y, and rewrite (5) as

(a - x)(a + x)/(a + y)(a - y) (y + b)/(x - b) (y - b)/(x + b) = 1.

which is just

(a - x)/(a - y) (b - y)/(b - x) = 1,
http://www.cut-the-knot.org/pythagoras/PPower.shtml


12/22

or

(a - x)/(a - y) = (b - x)/(b - y).

The latter is equivalent to

(a - b)(x - y) = 0.

Finally, except for the degenerate case of a = b where the chord becomes the tangent, we

get x = y.The latter is of course also true when a = b, i.e. when all the points on PQ coalesce

into one -- M.

Proof 10

I shall again prove Klamkin's generalization.

Consider twopencils of lines, one through A, the other through C. Relate those lines from

the two pencils that intersect on the given circle, such that AP corresponds CP, AD to CD, AB,

to CB, and AQ to CQ. All the pertinent angles between the pairs of related lines are equal as

subtending the same arc. Therefore, A(PDBQ) = C(PDBQ). By theproperties of cross-

ratio, A(PDBQ) = (PXWQ), whereasC(PDBQ) = (PZYQ), which gives

(PXWQ) = (PZYQ).

By definition,

(6) WP/WX : QP/QX = YP/YZ : QP/QZ.

But since ZM = MW, we also have WP = QZ, so that we can rewrite (6) as

(6') QX/WX = YP/YZ : QP/QZ.

Using the convention of the previous proof, (6') becomes

(a + x)/(b + x) = (a + y)/(b + y),

which is equivalent to (a - b)(x - y) = 0. Since a > b, the only possibility is x = y.
http://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtmlhttp://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml#lemmahttp://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml#remarkhttp://www.cut-the-knot.org/triangle/pythpar/Geometries.shtml#pencilhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#gb


13/22

Proof 11

The proof is due to Professor Hiroshi Haruki [Honsbergerand Diamonds, 136-138] and

depends on his wonderful lemma, which we apply to the diagram below:

Think of A and C as being two positions of a variable point traversing the circle.

Then Haruki's lemmaleads to

XPMQ / XM = MPYQ / YM,

which, because of MP = MQ, is simplified to

XP / XM = YQ / YM.

Adding 1 to both sides gives

(XP + XM) / XM = (YQ + YM) / YM.

Applying MP = MQ again, we obtain the required XM = YM.

Note that with a little change in the above we can easily obtainMorrey Klamkin'sgeneralization.

Proof 12

I have borrowed this proof from Paris Pamfilos' web site which since disappeared or moved.
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#gbhttp://www.cut-the-knot.org/Curriculum/Geometry/Haruki.shtmlhttp://www.cut-the-knot.org/Curriculum/Geometry/Haruki.shtmlhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Diamondshttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#H


14/22

Let S and T be the points of intersection of AD, BC and AC, BD. ThenST is thepolarof M

with respect to the given circle. In particular, this means that, if O is the center of the

circle, then OM is perpendicular to ST. Since M is the middle of PQ, PQ, too, is perpendicular

to OM. Hence, ST||PQ.

Further, from the constructionvia the complete quadrilateralof pairs of conjugate points,

it follows that the conjugate (F) of M with respect to the pair C, D lies on ST. As a

consequence, the lines SD, SC, SM, SF form a harmonic bundle. Any line crossessuch a

bundle in conjugate points. In particular, PQ crosses it in points X, Y, M, and infinity (recall

that ST||PQ), which me knowimplies that MX = MY.

Remark

Paris' proof actually shows more. For M not necessarily the midpoint of PQ, PQ will cross ST at

some point N say. As before there are several harmonically conjugate pairs. In particular,

(XYMN) = -1 and, of course, (PQMN) = -1. For these two,

1/XM + 1/YM = 2/NM, and1/PM + 1/QM = 2/NM,

where all segments are signed. Subtracting and reversing to regular lengths, we get Candy's

generalization, because, as signed segments, XM and YM and also PM and QM have different

signs.

Proof 13
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Candyhttp://www.cut-the-knot.org/pythagoras/ComplexHarmonicRatio.shtml#butterflyhttp://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml#infinityhttp://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtmlhttp://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml#harmonichttp://www.cut-the-knot.org/ctk/CompleteQuadrilateral.shtmlhttp://www.cut-the-knot.org/pythagoras/HarmonicRatio.shtmlhttp://www.cut-the-knot.org/Curriculum/Geometry/PolePolar.shtmlhttp://www.cut-the-knot.org/pythagoras/StraightEdgePolar.shtml


15/22

This one comes from the first volume of the encyclopedic Problems in Planimetryby V. V.

Prasolov (#2.54). We shall be proving Klamkin's generalization.

Reflect points A, B, D and X in the perpendicular to PQ through its midpoint M. Denote the

reflections A', B', D', and X'. A', B', D' lie on the given circle, whereas X' lies on PQ. We wish

to show that X' coincides with Y.

Since Z and W are symmetric in M, A'B' passes through W. Further, we may observe the

equality of several directed (or signed) anglesbetween pairs of straight lines:

(WY, CY) = (B'B, CB) (BB'||PQ)

(B'B, CB) = (B'A', CA') (subtended by arc B'C)

(B'A', CA') = (WA', CA') (A'B' passes through W)

Hence quadrilateral WYA'C is cyclic. We have an additional chain of equalities among angles:

(WX', A'X') = (DD', A'D') (DD'||PQ)

(DD', A'D') = (DC, A'C) (subtended by same arc)

(DC, A'C) = (WC, A'C) (W lies on CD)

Hence quadrilateral WX'A'C is also cyclic and has exactly same circumcircle as WYA'C. Since,

besides W, PQ intersects this circle in at most one other point, Y = X'. Q.E.D.

Proof 14

While pursuing a delightful proof by Professor Haruki, I have first overlooked what R.

Honsberger calls a standard proof [Honsbergerand Diamonds, 135-136]
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Diamondshttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Hhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Harukihttp://www.cut-the-knot.org/Curriculum/Geometry/InscribedAngles.shtml#directedAnglehttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#gbhttp://www.cut-the-knot.org/pythagoras/Butterfly.shtml#VP2


16/22

We prove the original problem, so that M is the center of the chord PQ which is thus

perpendicular to the diameter MO. Let B' be the reflection of B in that diameter. In

particular, BB' is also perpendicular to OM and is divided by OM into equal parts. BB'M is

isosceles and its base angles 1 and 2 are equal. Further, BB'||PQ. Angle 3 is alternatewith 1

and angle 4 with 2. It follows that all four are equal.

In a cyclic quadrilateralABB'D angles at B (2) and D (5) are supplementary. They add up to

180. If so, angles 3 and 5 also add up to 180, which makes quadrilateral MXDB' cyclic. In this

quadrilateral angles 6 and 7 subtend the same chord MX. They are thus equal. For the same

reason, but in the given circle, angles 7 and 8 are equal. Which shows that angles 6 and 8 are

also equal.

Now consider triangles MBY and MB'X. They have equal sides MB and MB', and equal pairs of

adjacent angles: 3/4 and 6/8. By SAS, the triangles are equal. Therefore, MX = MY.

Proof 15
http://www.cut-the-knot.org/pythagoras/EuclidRef_I.shtml#(I.4)http://www.cut-the-knot.org/Curriculum/Geometry/CircumQuadri.shtml#suphttp://www.cut-the-knot.org/pythagoras/Copernicus.shtml#cyclehttp://www.cut-the-knot.org/triangle/pythpar/AnglesInTriangle.shtml#nomenclature


17/22

This proof is by weininjieda from Yingkou, China who plans to become a teacher of

mathematics, Chinese and history. As inproof #8, weininjieda compares areas of several

triangles based, in particular, on the observation related to cyclic quadrilateral. For example,

quadrilateral ACDP is split by the diagonal AD into two triangles, APD and ACD whose opposing

angles at P and C add up to 180. Thus,

Area(APD)/Area(ACD) = APDP / ACCD.

Denote, as before, a = MP = MQ, x = MX, y = MY. Then

PX/MX MY/QY = (a - x)/x y/(a - y)

= Area(APD)/Area(AMD) Area(CMB)/Area(CQB)

= Area(APD)/Area(ACD) Area(CAD)/Area(CBD)

Area(BDC)/Area(BQC) Area(BMC)/Area(AMD)

= APDP/ACCD ADAC/BCBD BDCD/BQCQ BMCM/AMDM

= APDPADBMCM / BCBQCQAMDM

= DP/CQ AP/BQ AD/BC BM/DM CM/AM

= MP/CM AM/MQ DM/BM BM/DM CM/AM

= MP/MQ
http://www.cut-the-knot.org/pythagoras/Butterfly.shtml#Proof8


18/22

= 1.

It follows that (a - x)/x = (a - y)/y, or a/x - 1 = a/y - 1 so that x = y.

Proof 16

This proof is by Greg Markowsky.

Since M is the midpoint of PQ, the diameter of the circle through M is obv iously

perpendicular to PQ. Reflect A and D in that diameter to obtain points A' and D'. Observe that

points A' and B form apair of reflectionabout M; and so are points C and D'. By Greg's

theoremabout two such pairs, lines A'D' and BC meet on PQ, i.e., at Y. Since A'D' is the

reflection of AD in the diameter through M, MX = MY.

Proof 17

The proof has been communicated to me by Giles Gardam, a member of Australia's 2007 and

2008 IMO teams, and is due to his mentor, Ivan Guo, who won a gold medal at IMO 2004.
http://www.cut-the-knot.org/Curriculum/Geometry/RightTrianglesSameBase.shtml#explanationhttp://www.cut-the-knot.org/Curriculum/Geometry/RightTrianglesSameBase.shtml#explanation


19/22

An interactive illustration is available elsewhere.

Reflect B and C and the circle T in M (by reflect in a point I mean a dilation by factor -1 about

that point). B', C', and T' be the reflections of B, C and circle T, respectively. Note that as M

is the midpoint of PQ, T' passes through P and Q.

Bypower of pointM with respect to T, AMBM=CMDM.

Thus AMB'M=C'MDM, so by converse of power of a point, AB'C'D is a cyclic quadrilateral.

Let its circumcircle be T''.

We now consider the three radical axesof the three circles, which are PQ, AD and B'C', and

concur by the radical axes theorem. Thus B'C' intersects PQ at X. The reflections of BC and

PQ are B'C' and PQ, so considering their intersections, we have that X is the reflection of Y,

thus M is the midpoint of XY.

Proof 18

This proof is by B. Elsner aliasMathOMan. (Check his link to an amazing videoof Professor

Jean-Pierre Kahane engaging passers-by in the street, using the butterfly theorem as one of

the attractions.)
http://images.math.cnrs.fr/Un-mathematicien-dans-la-rue.htmlhttp://www.mathoman.com/index.php/1571-exercice-sur-les-cordes-d-un-cerclehttp://www.cut-the-knot.org/triangle/remarkable.shtml#radicalcenterhttp://www.cut-the-knot.org/triangle/remarkable.shtml#radicalhttp://www.cut-the-knot.org/pythagoras/Copernicus.shtml#cyclehttp://www.cut-the-knot.org/triangle/remarkable.shtml#powerhttp://www.cut-the-knot.org/Curriculum/Geometry/RadicalInButterfly.shtml


20/22

For a plain analytic proof, assume for simplicity that the radius of the circle is 1, PQ lies on

the horizontal x-axis, with the origin at M and the center of the circle O

at (0, m), |m| < 1. The equation of the circle then is x + (y - m) = 1 and that of PQ is y = 0.

We may assume that neither AB nor CD coincide with PQ. Since both path through the origin,

their equations are, say, x = ay, for AB, and x = by, for CD. Observe that in this form AB and

CD are allowed to be vertical. We may assume a b as otherwise the problem does not make

much sense.

Substituting x = ay in the equation of the circle, we see that the ordinates of the

points A(xA, yA) andB(xB, yB) satisfy the equation ay + (y - m) = 1, implying

(1 + a)y - 2my + (m - 1) = (1 + a)(y - yA)(y - yB).

A similar relation holds for the ordinates of points C, D. Comparing the coefficients, we see

that

(7)

(1 + a)(yA+ yB) = 2m, (1 + a)yAyB= m - 1,

(1 + b)(yC+ yD) = 2m, (1 + b)yCyD= m - 1.

Points X and Y lie on the x-axis so that their ordinates vanish: X(xX, 0) and Y(xY, 0). X is

collinear with A and D, implying

(xX- xD) / (yX- yD) = (xA- xD) / (yA- yD)

In other words,


21/22

(xX- xD)(yA- yD) = (yX- yD)(xA- xD),

or, substituting xA= a yAand xD= b yD,

(8) xX(yA- yD) = (a - b) yAyD.

Similarly,

(8') xY(yB- yC) = (a - b) yByC.

Multiplying the first of these equations by (yB- yC) and the second by (yA- yD) and adding the

products shows (after not difficult algebraic manipulations using (7)) that

(yA- yD)(yB- yC)(xX+ xY) = 0.

Observe, that, under the assumption a b, (8) and (8') imply yA yDand yB yCso that we

may conclude that xX+ xY= 0, which immediately gives MX = MY.

Proof 19

Mikhail Goldenberg (The Ingenuity Project, Baltimore, MD) and Mark Kaplan (Towson State

University) came up with a purely projective proof of the Butterfly theorem

I placed the proof on a separate page.
http://www.cut-the-knot.org/pythagoras/ProjectiveButterfly.shtmlhttp://www.ingenuityproject.org/


22/22

Proof 20

Another projective proof is due to Hubert Shutrick and is based on the properties

ofprojective involution:

The pencil of conics that pass through A, B, C, D in the diagram include the circle and

the pair of lines AB, CD, whose intersection H is a double point of the involution on

PQ and the other double point must be the point at infinity since H is the midpoint of

PQ. Hence, it is also the midpoint of XY because ADBC is another degenerate conic in

the pencil.

Reference

1. H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, MAA, 1967

2. Alex D D Craik and John J O'Connor, Some unknown documents associated with

William Wallace (1768-1843), BSHM Bulletin: Journal of the British Society for the

History of Mathematics, 26:1, 17-28

3. S. L. Greitzer,Arbelos, v 5, ch 2, pp 38-39, MAA, 1991

4. R. Honsberger, The Butterfly Problem and Other Delicacies from the Noble Art of

Euclidean Geometry I, TYCMJ, 14 (1983), pp. 2-7.

5. R. Honsberger,Mathematical Diamonds, MAA, 2003

6. V. V. Prasolov, Essays On Numbers And Figures, AMS, 2000

7. V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986, in Russian

8. V. V. Prasolov, Problems in Planimetry, v 2, Nauka, Moscow, 1986, in Russian

9. D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of

Elementary Mathematics, v 2, Moscow, 1952.
http://www.amazon.com/exec/obidos/ISBN=0821819445/ctksoftwareincA/http://www.amazon.com/exec/obidos/ISBN=0883853329/ctksoftwareincA/http://pdfserve.informaworld.com/196860__932897563.pdfhttp://www.amazon.com/exec/obidos/ISBN=0883856190/ctksoftwareincA/http://www.cut-the-knot.org/wiki-math/index.php?n=Geometry.Involution

butterfly theorem 1

Documents