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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:
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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
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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
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= 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
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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.
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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
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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
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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.
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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)
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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.
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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,
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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.
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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.
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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
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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]
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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
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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
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= 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.
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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.shtml8/12/2019 Butterfly Theorem 1
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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,
8/12/2019 Butterfly Theorem 1
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(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/8/12/2019 Butterfly Theorem 1
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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.InvolutionRecommended