Unpublished notes - Geometry
Don't expect too much from these notes. They were most often written for one specific purpose (such as answering a mathematical question on a newsgroup or a forum) and are anything but well-written. They consist by far not only of new and original results.
Darij Grinberg, Radical axes revisited
(version 7 September 2007).
PDF file.
This note is devoted to some not so commonly known
properties of radical axes in geometry. We start with the
following fact:
Theorem 1 (Casey). Let A and B be two
distinct points, and let x and y be two numbers. Let P be
a point. Let A(x) mean the circle with center A and
radius x, and similarly define B(y).
The two lines PQ and AB are parallel. If we direct these
lines in the same way (that means, we direct them in such
a way that equal vectors along these lines correspond to
equal directed segments), then
(power of P with respect to circle A(x)) - (power of P
with respect to circle B(y)) = 2 * (directed distance
from P to radical axis of circles A(x) and B(y)) * AB.
Among the consequences of this fact is:
Theorem 2. Let A, B, C be three points
on a line g, and x, y, z be three numbers. Let the
radical axis of the circles B(y) and C(z) meet g at X,
and similarly define Y and Z. Then,
YZ = 1/2 * (x˛ / (AB * AC) + y˛ / (BC * BA) + z˛ / (CA
* CB) - 1) * BC
and YZ / BC = ZX / CA = XY / AB, where the segments are
directed.
This immediately yields a criterion for coaxality of
three circles (by setting YZ = 0), which - in the case
when the circles meet at two points - yields the
following consequence:
Theorem 4. Let A, B, C be three pairwise
distinct points on a line g, and let P be a point. Then,
AP˛ / (AB * AC) + BP˛ / (BC * BA) + CP˛ / (CA * CB) =
1
with directed segments.
This is an equivalent version of Stewart's theorem, but
one that can be more easily memorized than the standard
version.
Another consequence of Theorem 2 is a construction of
radical axes using inversion (Theorem 6), which, in turn,
yields a characterization of the center of the Taylor
circle of a triangle as a radical center.
Darij Grinberg, An unexpected
application of the Gergonne-Euler theorem
(version 22 March 2007).
PDF file.
This note solves a problem from the IMO longlist 1976
proposed by Great Britain:
Let ABC and A'B'C' be two triangles on a plane. Let the
lines BC and B'C' intersect at X, and similarly define
two points Y and Z. Let the parallel to BC through A
intersect the parallel to B'C' through A' at X', and
similarly define two points Y' and Z'. Prove that the
lines XX', YY', ZZ' concur at one point.
The proof uses the Gergonne-Euler theorem about the
ratios in which concurrent cevians of a triangle divide
each other. As a sidenote, a simple equivalence between
this Gergonne-Euler theorem and the van Aubel theorem is
established.
Darij Grinberg, On cyclic quadrilaterals
and the butterfly theorem
(version 16 February 2007).
Zipped PDF file.
The note begins with a slight extension of what is
usually referred to as the "butterfly theorem":
Theorem 1. Let k be a circle with center
O, and let A, B, C, D be four points on this circle k.
Let AC and BD intersect at P. Let g be a line through the
point P such that P is the orthogonal projection of the
point O on this line g. Let the line g intersect AB and
CD at X and Z. Then, the point P is the midpoint of the
segment XZ.
We are giving two proofs of this fact. The first one
applies the Pascal theorem to establish a generalization
by Klamkin. The second one combines an affine theorem,
which can be shown using Ceva and Desargues, with some
properties of cyclic quadrilaterals. Finally, an
application of the first proof is given - a triangle
geometry problem from the St. Petersburg Mathematical
Olympiad 2002.
Support page: List of clickable
references.
Darij Grinberg, Isogonal conjugation with
respect to a triangle (version
23 September 2006).
Zipped PDF file; a possibly outdated version
(not ZIPped) is also downloadable from the MathLinks
forum as an attachment in the topic
"Isogonal conjugation with respect to a
triangle".
This note gives a detailed introduction into a part of
triangle geometry, and gives new or simplified proofs of
a few recent results.
First, the notions of isogonal lines or isogonals, and of
isogonal conjugates with respect to a triangle are
introduced. Proofs are given for the existence of the
isogonal conjugate, for the fact that the isogonal
conjugate of the circumcircle is the line at infinity,
for the pedal circle theorem and for the relation between
the isogonal conjugate and reflections in sidelines, as
well as for other properties.
Then, the following result of Antreas P. Hatzipolakis and
Paul Yiu is shown:
Let P be a point in the plane of a triangle ABC. The
lines AP, BP, CP intersect the lines BC, CA, AB at the
points A', B', C'. Let Q be the isogonal conjugate of the
point P wrt the triangle ABC. Then, the reflections of
the lines AQ, BQ, CQ in the lines B'C', C'A', A'B' concur
at one point.
The proof of this result given in this note is a
simplification of an argument by Jean-Pierre Ehrmann (see
the
corresponding MathLinks topic).
Then, an observation of José Carlos Chávez Sandoval (again
first published on the MathLinks forum) is proven:
Let P be an Euclidean point in the plane of a triangle
ABC, and let D, E, F be the reflections of this point P
in the perpendicular bisectors of the segments BC, CA,
AB. Denote by AM, BM, CM
the midpoints of the segments BC, CA, AB, and by DM, EM,
FM the midpoints of the
segments EF, FD, DE. Let Q be the isogonal conjugate of
the point P wrt the triangle ABC, and let Q' be the
complement of the point Q wrt the triangle ABC. Then, the
lines AMDM,
BMEM,
CMFM
pass through the point Q'.
Next, the isogonal conjugates of the incenter, the
excenters, the orthocenter, the circumcenter and the
centroid of the triangle are identified, and a short
introduction into the notion of antiparallels is given.
This part is very easy and well-known, but I have not
often seen it written up with proofs, so this is my
attempt at it.
Finally, we show a fact which is not new but seems to be
underused:
Let ABC be a triangle, and let P and Q be two points on
the perpendicular bisector of the segment BC. Then, the
four equations < ABP = < ACQ, < ACP = < ABQ,
< BCP + < BCQ = < BAC and < PBC + < QBC =
< BAC (where all angles are directed angles modulo
180°) are pairwisely equivalent, and they are all
equivalent to the assertion that the points P and Q are
inverse to each other wrt the circumcircle of triangle
ABC. Besides, if this assertion holds, then the lines AP
and AQ are isogonal to each other wrt the angle CAB.
This is used to prove a simple property of the isogonal
conjugates of Kiepert points.
Darij Grinberg, Circumscribed
quadrilaterals revisited
(version 13 September 2008).
PDF file.
The aim of this note is to prove some new properties of
circumscribed quadrilaterals and give new proofs to
classical ones.
Among others, the result from my note "A theorem on circumscribed
quadrilaterals" receives a new proof, but also
some additional assertions and metrical identites are
established. A synthetic proof is given to an
identity from the China TST 2003: In a circumscribed
quadrilateral ABCD with incenter O, we have OA * OC + OB
* OD = sqrt( AB * BC * CD * DA ). Formulae for the area
and the inradius of a circumscribed quadrilateral are
established. A
problem from the final round of the All-Russian
Mathematical Olympiad 2005 is solved: The incenter O
of a circumscribed quadrilateral ABCD coincides with its
centroid if and only if OA * OC = OB * OD.
Darij Grinberg, An adventitious angle
problem concerning sqrt(2) and pi/7.
PDF file.
In this note, I present two synthetic solutions - one by
Stefan V., one apparently original - to the following
problem, which arose
in a MathLinks discussion:
Let ABC be an isosceles triangle with AB = AC and BC = 1.
Let P be a point on the side AB of this triangle which
satisfies AP = 1.
Prove that CP = sqrt(2) holds if and only if < CAB =
pi/7.
Darij Grinberg, Three properties of
the symmedian point.
PDF file.
This note provides synthetic proofs of three results
concerning the symmedian point of a triangle:
1 (Theorem 6, a part of problem G5 from the IMO Shortlist
2000). Let L be the symmedian point of a
triangle ABC. The symmedians BL and CL of triangle ABC
intersect the sides CA and AB at the points E' and F',
respectively. Denote by E'' and F'' the midpoints of the
segments BE' and CF'. Then:
a) We have < BCE'' = - < CBF''
(where we use directed angles modulo 180°).
b) The lines BF'' and CE'' are symmetric
to each other with respect to the perpendicular bisector
of the segment BC.
2 (Theorem 8, originated from a locus problem by
Antreas P. Hatzipolakis). Let ABC be a triangle,
and let A' be the midpoint of its side BC. Let S be the
centroid and L the symmedian point of triangle ABC. Let J
be the point on the line SL which divides the segment SL
in the ratio SJ / JL = 2 / 3.
Let X be the point of intersection of the median AS of
triangle ABC with the circumcircle of triangle ABC
(different from A). Denote by U the orthogonal projection
of the point X on the line BC, and denote by U' the
reflection of this point U in the point X.
Then, the line AU' passes through the point J and bisects
the segment LA'.
3 (Theorem 10, a classical result). Let L be the
symmedian point of a triangle ABC. Let the symmedian AL
of triangle ABC meet the side BC at a point D' and the
circumcircle of triangle ABC at a point X' (different
from A). Then, (AL / LD') / (AX' / X'D') = - 1/2.
Darij Grinberg, On the paracevian
perspector.
Zipped PDF file.
In this note, I provide a synthetic proof and a corollary
of a theorem found by Eric Danneels (Hyacinthos message
#10135):
Let P and Q be two points in the plane of a triangle ABC.
The parallels to the lines AP, BP, CP through the point Q
intersect the lines BC, CA, AB at the points U, V, W.
The parallels to the lines BC, CA, AB through the point Q
intersect the lines AP, BP, CP at the points U', V', W'.
Then, the lines UU', VV', WW' concur at one point.
Darij Grinberg,
The Mitten
point as radical center.
Zipped PS file.
The Mitten point of a triangle, defined as the perspector
of the medial and excentral triangles, is shown to be the
radical center of a variable circle triad. In fact, if
the sidelines BC, CA, AB of a triangle ABC are directed
such that the segments BC, CA, AB are positive, and Ab, Ac,
Bc, Ba,
Ca, Cb
are points on the lines CA, AB, AB, BC, BC, CA,
respectively, fulfilling BBa
= CaC = CCb
= AbA = AAc
= BcB, then the pairwise
radical axes of the circles ABcCb, BCaAc, CAbBa are the lines IaMa,
IbMb,
IcMc,
where Ia, Ib, Ic
are the excenters of triangle ABC and Ma,
Mb, Mc
are the midpoints of BC, CA, AB. The radical center of
the three circles is the Mitten point of triangle ABC.
Darij Grinberg, A tour around
Quadrilateral Geometry.
Zipped PDF file.
The tour begins with an arbitrary quadrilateral ABCD. The
external angle bisectors of its angles enclose another
quadrilateral XYZW, which is shown to be inscribed. Its
circumcenter M turns out to be equidistant from the
perpendiculars to the sidelines of ABCD through the
vertices of XYZW.
Next, we pay attention to the case of a circumscribed
quadrilateral ABCD. The incenter O of ABCD happens to
coincide with the intersection of XZ and YW, and the
midpoint of the segment OM' is equidistant from the
perpendicular bisectors of the sides of ABCD. The proof
idea for this result originates from Marcello Tarquini.
Finally, let ABCD be an arbitrary quadrilateral again. We
consider the quadrilateral A'B'C'D' formed by the
pairwise intersections of the perpendicular bisectors of
adjacent sides of ABCD. This quadrilateral A'B'C'D' is
called the PB-quadrilateral of ABCD.
The above implies that the PB-quadrilateral of a
circumscribed quadrilateral is circumscribed, too.
The tour ends with the proof of a converse of this fact:
If ABCD is an arbitrary, but not inscribed,
quadrilateral, and the PB-quadrilateral A'B'C'D' is
circumscribed, then ABCD is circumscribed, too.
An easy-to-prove lemma states that any quadrilateral is
homothetic to the PB-quadrilateral of its
PB-quadrilateral.
Darij Grinberg, Generalization of the
Feuerbach point.
Zipped PDF file.
Using directed angles modulo 180° (crosses), I prove a
suite of theorems about pedal circles and orthopoles of
lines through the circumcenter of a triangle.
Given a triangle ABC, the midpoints A', B', C' of the
sides BC, CA, AB form the medial triangle A'B'C' of
triangle ABC. If U is the circumcenter of ABC and P is
any point different from U, then the reflections x, y, z
of the line PU in the sidelines of the medial triangle
A'B'C' concur at one point L lying on the nine-point
circle of triangle ABC. The reflections X', Y', Z' of L
in B'C', C'A', A'B' are the feet of the perpendiculars
from A, B, C to PU, and the point L itself is the
orthopole of PU with respect to triangle ABC. Let XYZ is
the pedal triangle of P with respect to triangle ABC;
then L lies on the circumcircle of triangle XYZ. If B'C'
and YZ meet at A", and B", C" are defined
cyclically, then L lies on the lines XA", YB",
ZC". (These results cover the first two Fontene
theorems.) Moreover, if HaHbHc
is the orthic triangle of ABC, then the lines
corresponding to PU in the triangles AHbHc, HaBHc, HaHbC pass through L, too. The
orthocenters D, E, F of triangles AYZ, BZX, CXY coincide
with the points corresponding to P in triangles AHbHc,
HaBHc,
HaHbC.
The points X, E, F, Ha
and L lie on one circle. This and more is established in
the first part of the paper.
In the second part, I consider the case when P is the
orthocenter H of triangle ABC. Then, the point L is the
meet of the Euler lines of triangles AHbHc, HaBHc, HaHbC. It is also shown that the
longest of the segments HaL,
HbL, HcL
equals the sum of the two others, solving a problem of
Victor Thebault in the "American Mathematical
Monthly".
In the third part, I regard the case when P is the
incenter O of triangle ABC. The points X, Y, Z are the
points where the incircle of triangle ABC touches the
sides BC, CA, AB. It is shown that the nine-point circle
of ABC touches the incircle, and the point of tangency is
L. This way of establishing the Feuerbach theorem is
presumably new. The point L is known as the Feuerbach
point of triangle ABC. The line OU being called the
diacentral line of triangle ABC, I show that L lies on
the diacentral lines of triangles AHbHc, HaBHc, HaHbC. If Am,
Bm, Cm
are the midpoints of AO, BO, CO, then L lies on the
circles with diameters XAm,
YBm, ZCm
(Michel Garitte) and on the nine-point circles of
triangles BOC, COA, AOB. Moreover, the reflections of the
line OU in the sidelines of triangle XYZ meet at L. The
orthocenters D, E, F of triangles AYZ, BZX, CXY coincide
with the incenters of triangles AHbHc, HaBHc, HaHbC.
Finally, in the last part of the paper, I return to the
original general case with an arbitrary point P and show
that the angle between the nine-point circle and the
pedal circle of P (this is the circumcircle of the pedal
triangle XYZ of P) is 90° - PBC - PCA - PAB. This
theorem is due to V. Ramaswamy Aiyer and generalizes the
Feuerbach theorem.
Some lemmas from my note "Anti-Steiner points with
respect to a triangle" are used in the proofs.
Darij Grinberg, Anti-Steiner points
with respect to a triangle.
Zipped PDF file.
Using directed angles modulo 180° (crosses), we prove
two results of S. N. Collings:
1. The reflections of a line g passing
through the orthocenter H of a triangle ABC in the
sidelines BC, CA, AB concur at a point on the
circumcircle of ABC.
2. This point also lies on the
circumcircles of triangles AYZ, BZX, CXY, where X, Y, Z
are the reflections of an arbitrary point P lying on g in
BC, CA, AB.
This point where the reflections of g in BC, CA, AB
concur is called the Anti-Steiner point of g
with respect to ABC.
Darij Grinberg, New
Proof of the Symmedian Point to be the centroid of its
pedal triangle, and the Converse.
Zipped PS file. See also New Proof
of the Symmedian Point to be the centroid of its pedal
triangle, and the Converse
for a zipped PDF file.
Numerous discussions in the Hyacinthos newsgroup
(beginning with Clark Kimberling's message #1) were
related to the fact that
a) the symmedian point of a triangle is
the centroid of its pedal triangle,
b) and it is the sole point having this
property.
In this note, I present an apparently new proof of part a)
(using the tangential triangle); furthermore, I extend a
standard proof of part a) (given, e. g.,
in Honsberger's book) to a proof of b).
The note is also are avaliable through the Hyacinthos
Files directory, as the file SymmedianPedal.zip (which
contains a Zipped PS file). This note has been announced
in Hyacinthos message #6237.
Darij Grinberg, On the Taylor center of a
triangle.
Zipped PS file.
I show a result which is not new but perhaps the first
time proven elementarily:
The center of the Taylor circle of a triangle is the
radical center of the circles centered at the vertices
and each of them touching the opposite sideline.
Darij Grinberg, Begonia points and coaxal
circles.
Zipped PS file.
This paper gives a synthetic proof of the following
theorem (Jean-Pierre Ehrmann, Hyacinthos message #7999):
Given a triangle ABC and a point P. Let A'B'C' be the
cevian triangle of P, and X, Y, Z the reflections of P in
the lines B'C', C'A', A'B'. Then the lines AX, BY, CZ are
concurrent.
This fact was called "Begonia theorem".
Jean-Pierre Ehrmann has given a shorter proof in
Hyacinthos message #8039.
My proof uses a result from circle geometry which seems
to be new:
Let ABC be a triangle and P a point. Let A' be the point
of intersection of the circle BCP and the line AP
different from P, and similarly define B' and C'. Let X,
Y, Z be the centers of the circles B'C'P, C'A'P, A'B'P.
Then, the circles PAX, PBY, PCZ are coaxal, i. e. they
have a common point different from P (or touch at P).
This latter fact is established with the help of the
Desargues Theorem.
Darij Grinberg, The
Lamoen Theorem on the Cross-Triangle.
Zipped PS file.
In 1997, Floor van Lamoen discovered a very nice and
useful theorem about perspective triangles and their
cross-triangles. I give a proof of this result by the
Desargues Theorem and also show some corollaries.
Darij Grinberg, The
Theorem on the Six Pedals.
Zipped PS file.
With the abbreviation "pedal" for "foot of
the perpendicular", the Theorem on the Six Pedals
states:
Let ABC be a triangle and P and Q two points. We
construct the perpendiculars from Q to the lines BC, CA,
AB. Let X, Y, Z be the pedals of the point P on these
perpendiculars. On the other hand, let X', Y', Z' be the
pedals of the point Q on the lines AP, BP, CP. Then, the
lines XX', YY', ZZ' concur.
I prove this (probably new) fact with the help of the
Ceva theorem in its trigonometric form.
Darij Grinberg, From the
Complete Quadrilateral to the Droz-Farny Theorem.
Zipped PS file. See also From the
Complete Quadrilateral to the Droz-Farny Theorem for a zipped PDF file.
A generalization of the Steiner-Miquel theorem on the
complete quadrilateral proven by Nikolaos Dergiades is
used to establish a remarkable result about triangles.
The latter faciliates the proof of the following fact:
Let g and g' be two mutually orthogonal lines through the
orthocenter H of a triangle ABC. The line g meets the
sidelines BC, CA, AB at A', B', C'; the line g' meets the
sidelines BC, CA, AB at A'', B'', C''.
a) The circles with diameters A'A'',
B'B'', C'C'' pass through the point H. (Nearly trivial)
b) These circles have a common point Q
different from H; this point Q lies on the circumcircle
of triangle ABC.
c) The circles with diameters A'A'',
B'B'', C'C'' are coaxal.
d) The midpoints of segments A'A'',
B'B'', C'C'' lie on one line. (A. Droz-Farny)
e) The equation B'C' : C'A' : A'B' =
B''C'' : C''A'' : A''B'' holds. (F. van Lamoen)
Darij Grinberg, The
Lamoen circle.
Zipped PS file.
Floor van Lamoen discovered the theorem that the
circumcenters of the 6 triangles in which a triangle is
subdivided by its medians are concyclic. Several proofs
are known; in this note, I give another proof (by
trigonometry).
The ZIP file equals the file Lamoen.zip in the Files
directory of the "Hyacinthos" newsgroup.
Support page: Proof of Theorem 2.
Darij Grinberg, Variations of the
Steinbart Theorem.
Zipped PS file. See also Variations
of the Steinbart Theorem for a zipped PDF file.
At first, I establish the following theorem, which was
partially shown by Oliver Funck and Stanley Rabinowitz:
Let ABC be a triangle and A'B'C' its tangential triangle.
Further, let A'', B'', C'' be any three points on the
circumcircle of triangle ABC. The lines A'A'', B'B'',
C'C'' concur if and only if either the lines AA'', BB''
and CC'' concur or the points AA'' /\ BC, BB'' /\ CA,
CC'' /\ AB are collinear. Here, the sign "/\"
means "intersection".
After proving this result, a property of the incircle
stated by Jean-Pierre Ehrmann in Hyacinthos message #6966
is derived from this theorem using poles and polars with
respect to circles.
Here
is the link to reference [1].
Darij Grinberg, On the feet of the
incenter on the perpendicular bisectors.
Zipped PS file.
We consider the perpendiculars OX, OY, OZ from the
incenter O of a triangle ABC to the perpendicular
bisectors of its sides BC, CA, AB. It is shown that
triangle XYZ is oppositely similar to triangle ABC, that
the lines AX, BY, CZ pass through the Nagel point of
triangle ABC, and that one of the segments OX, OY, OZ is
equal to the sum of two others.
Darij Grinberg, A Bundeswettbewerb
Mathematik problem and its relation to the Nagel point of
a triangle.
Zipped PS file. See also A
Bundeswettbewerb Mathematik problem and its relation to
the Nagel point of a triangle for a zipped PDF
file.
A problem in the Bundeswettbewerb Mathematik 2003 asked
to prove the following property of triangles:
In a parallelogram ABCD, points M and N are chosen on the
sides AB and BC in a such way that they don't coincide
with a vertex, and that the segments AM and NC have equal
length. Let Q be the intersection of the segments AN and
CM. Then, DQ bisects the angle ADC.
This problem is solved and used to derive the Nagel
theorem (that the incenter of a triangle is the Nagel
point of the medial triangle).
German version: Eine Aufgabe
aus dem Bundeswettbewerb Mathematik und der Nagelsche
Punkt eines Dreiecks at http://www.dynageo.de/discus/messages/5/112.html.
Darij Grinberg, A theorem on
circumscribed quadrilaterals.
Zipped PS file.
Let ABCD be a circumscribed quadrilateral with incenter
O. The perpendicular to AB through A meets BO at M. The
perpendicular to AD through A meets DO at N. Then MN is
perpendicular to AC. I give two proofs for this result,
including an ingenious synthetic proof by Nikolaos
Dergiades. A degenerate case is also discussed.
Darij Grinberg, On the Lemoine
circumcevian triangle.
Zipped PS file.
A new proof is given for the fact that the symmedian
point of a triangle is the symmedian point of its
circumcevian triangle.
Darij Grinberg, New
Insight on the Ninepoint Circle.
The note is on a new proof that the midpoints of the
sides, the feet of the altitudes and the midpoints
between the vertices and the orthocenter of a triangle
lie on one circle.
This a corrected version of the contents of the
file NewNinepoint.zip in the Files directory of the "Hyacinthos" newsgroup.
Support page: Two messages in the
"Hyacinthos" newsgroup related to the note.
German version: Neuer Zugang zum
Feuerbachkreis at http://www.dynageo.de/discus/messages/5/112.html.
Darij Grinberg, Synthetic
proof of Paul Yiu's excircles theorem.
Zipped PS file. See also Synthetic proof
of Paul Yiu's excircles theorem for a zipped PDF file.
In this note, I give a synthetic proof for Paul Yiu's
excircles theorem, which states that, if ABC is a
triangle with orthocenter H, then the triangle whose
sides are the the polars of A, B, C with respect to the
A-excircle, B-excircle, C-excircle of triangle ABC,
respectively, is perspective to triangle ABC, and the
perspector is H.
An old GIF version of the note can be downloaded from the
Files directory of the
"Hyacinthos" newsgroup under YiuSynth.zip (announced in Hyacinthos
message #6176).
Darij Grinberg, Problem:
The ortho-intercepts line.
Zipped PS file.
In this note, I present Jacques Hadamard's proof of a
theorem about triangles and orthogonal lines. The proof
is an elegant demonstration of the application of
polarity in geometry.
See also: Bernard Gibert, Orthocorrespondence and Orthopivotal Cubics, Forum Geometricorum 3 (2003) pages 1-27.
The ZIP file equals to the file OrthoIntercept.zip in the
Files directory of the "Hyacinthos" newsgroup.
Darij Grinberg, The Euler
point of a cyclic quadrilateral.
Zipped PS file.
This note gives proofs of some results about cyclic
quadrilateral (mostly old ones). These results concern
the Euler point of the cyclic quadrilateral (coinciding
with the so-called anticenter).
The ZIP file equals the file EulerQuad.zip in the Files
directory of the "Hyacinthos" newsgroup.
Darij Grinberg, A new proof of the
Ceva Theorem.
Zipped PS file.
I present a possibly new proof of the Ceva theorem. The
proof makes use of auxiliary parallels (namely, parallels
to the sides of the triangle through the point).
The same proof in a similar description can be found at my
geometry-college message "New Proof of Ceva's
Theorem", which equals to "Hyacinthos"
message #6683.
I have been regularily posting in the "Hyacinthos" and geometry-college newsgroups (starting with 1 Jul 2002) and on the MathLinks forum.
I have set up a Schröder points database for the Schröder points and related problems.
Unpublished notes - Geometry
Darij Grinberg