Spatial Analogues of Ceva’s Theorem and its Applications
Nadav Goldberg
12020 Montrose Village Terrace
Rockville, MD 20852 USA
nadav2@hotmail.com
Abstract
Some interesting and useful theorems of planar geometry have interesting analogues
in three-dimensional geometry as well. Within the framework of this project, such an
analogue for Ceva’s theorem was found. It was then used to solve problems in threedimensional space that correspond to planar corollaries of the original Ceva’s theorem.
Introduction
Many important planar theorems have solid geometric analogues. For example, the
famous Pythagorean theorem has many analogues in solid geometry. Other lesser-known but
important theorems also have analogues. These analogies between theorems on the plane and
theorems in space are not only elegant, but are many times quite useful. If a theorem on the
plane is useful in problem solving, then perhaps the spatial analogue may be used to formulate
and solve similar problems in space.
A concrete example, which will be used later, can be provided. There is a theorem
that an angle bisector in a triangle divides the opposite side into lengths proportional to the
adjacent sides. A proof of the stereometric analogue is presented here.
Theorem 1 (Dihedral angle bisector theorem)
Let ABCD be a tetrahedron (see figure 1). Point
X is on BC and plane AXD is the dihedral angle
bisector of dihedral angle ( ADB, ADC ) . Then
BX S ABD

.
XC S ACD
A
V
The proof is not long. The ratio ABXD is
C
V ACXD
X
S BXD
equal
to
because both tetrahedrons have a
Figure 1
B
S CXD
BX
common height from A to (CBD). The ratio of areas can also be simplified further to
XC
because both triangles have a common height from D to BC . Now, the original ratio of
volumes can be rewritten in another way. The height from X to (ABD) is equal to the height
from X to (ACD). This is a property of the dihedral angle bisector. This means that in the
ratio of volumes, the height cancels out from numerator and denominator, leaving the ratio of
areas of triangle ABD to triangle ACD. To recapitulate,
D
S ABD V ABXD AX
. Q.E.D.


S ACD V ACXD XD
This is an elegant example of an extension of a planar theorem to space. Whereas on
the plane the angle bisector divides the opposite side into lengths proportional to the lengths
of the adjacent sides, in space, the dihedral angle bisector divides the opposite edge into
lengths proportional to the areas of the adjacent faces. It is a very direct analogy.
An aim of this project was to extend two other useful planar theorems to space: the
theorem of Menelaus and Ceva’s theorem. Before learning about these two closely connected
statements, one should know a bit about their discoverers.
Giovanni Ceva
Giovanni Ceva was born in 1647 and learned at a Jesuit college in his hometown of
Milan. He proceeded to study at the University of Pisa and began teaching there until his
appointment as professor of mathematics at the University of Mantua in 1686 for the
remainder of his life. Ceva supported the local rulers until the area was taken over by Austria,
which he then quickly moved to support. Ceva’s great discoveries were mostly in the field of
geometry. In his 1678 work, De lineis rectis, he first published an important geometric
theorem overlooked by the Greeks that is now named “Ceva’s theorem.” He also
rediscovered and published the theorem of Menelaus. In his other works, he applied
geometry to mechanics and hydraulics and even anticipated calculus to some extent. His
1711 work, De Re Nummeraria, was one of the first works in mathematical economics*.
Theorem 2 (Ceva’s theorem)
Let ABC be a triangle (see figure 2) with point X
on AB , Y on BC , and Z on CA . AY , BZ , and
CX are concurrent if and only if
B
Y
X
A
C
Z
Figure 2
AX BY CZ


 1.
XB YC ZA
For proofs of this well-known statement, the reader is referred to [1] or [4]. Ceva’s
theorem provides an easy and useful tool for proving concurrency of segments, such as in the
following famous corollary, which will be later used.
Theorem 3 (Gergonne point)
Let a circle be inscribed in triangle. The segments drawn from each vertex of the
triangle to the point of tangency on the opposite side are concurrent.
This theorem may be easily proven with the aid of Ceva’s theorem. For specifics, see
[4].
*
The biographical information presented here is from [6].
Menelaus of Alexandria
The mathematician Menelaus was born around 70 AD in Alexandria. He lived in
Rome for a time, as evidenced by a description by Plutarch of a conversation he had with the
philosopher Lucius in Rome. Both Pappus and Proclus mention Menelaus and there is a
reference to his astronomical observations in the records of Ptolemy. Tenth century Arab
records indicate that he wrote a great number of books. Unfortunately, only one survives and
it has been heavily edited by Arabic authors. His work, Sphaerica, deals with spherical
triangles and their application to astronomy. It was the first detailed exposition of spherical
geometry. In this work, he proved a spherical version of the planar theorem of Menelaus.
Interestingly, the planar version was well known already*.
Theorem 4 (Theorem of Menelaus)
Let ABC be a triangle (see figure 3) with point
B
X
X on AB , Y on BC , and Z on CA . X, Y, and Z
are collinear if and only if
Y
Z
A
Figure 3
AX BY CZ


 1.
XB YC ZA
The exact proof may be found in [1] or [3]. Notice that both the theorems of Ceva and
Menelaus have the same equation. Furthermore, they speak of similar but opposte concepts.
Ceva’s theorem relates to the concurrentness of segments in a triangle, and Menelaus’s
theorem relates to collinearity of points on a triangle. This idea of similarity between the two
theorems is called “duality.” In fact, the two theorems may even be shown to be equivalent:
one may use one to prove the other [4].
Theorem of Menelaus in Space
First attempts to formulate an analogue of this theorem by examining the ratios of
areas when a plane passes through a tetrahedron failed. While doing background research on
solid geometry, however, an interesting three-dimensional extension of the theorem of
Menelaus was found in [2].
D
W
Theorem 5
Let ABCD be a space quadrilateral (see figure 4).
Point X lies on AB , Y lies on BC , Z is on CD , and W is
on DA . Points X, Y, Z, and W are coplanar if and only
if
AX BY CZ DW



 1.
C
XB YC ZD WA
Z
A
X
Y
This theorem is a direct analogue of the planar
original. Sides of a triangle become edges of a space
quadrilateral and a line intersecting the triangle becomes
a plane intersecting the space quadrilateral. The equation is still nearly identical. This
extension, like the original, uses ratios of distances. If one wishes to make the extension more
“three-dimensional,” then the ratios of distances can be replaced with ratios of areas by
Figure 4
*
B
This information was taken from [5].
multiplying the numerator and denominator of each ratio by one-half a height common to the
two bases.
Results
Just as Ceva’s theorem on the plane has a duality with the theorem of Menelaus, it was
hoped that there would be a similar duality in the spatial analogues. This was the one of the
guiding principles in the development of a three-dimensional Ceva’s theorem. Such a duality
was found. Instead of a plane cutting through the edges of a quadrilateral at four points, four
planes are constructed from each edge to to the opposite one. In both cases, the equation is
the same and quite similar to the planar version.
D
W
Z
P
A
C
X
Theorem 6 (Ceva’s theorem in
space)
Let
ABCD
be
a
space
quadrilateral (see figure 5).
Point X lies on AB , Y lies on BC ,
Z is on CD , and W is on DA .
Four planes AZB, BWC, CXD, and
DYA intersect at exactly one point
if and only if
(1)
Y
AX BY CZ DW



1
XB YC ZD WA
Figure 5
B
In other words, four planes drawn from the edges of a quadrilateral to a point on the
opposite edge intersect at exactly one point if and only if equation (1) is true.
Let us introduce this theorem in symbolic form.
D
W
C
T
P
Given:
 Space quadrilateral ABCD
 X  AB , Y  BC , Z  CD , W  DA
 (AZB) ∩ (BWC) ∩ (CXD) ∩ (DYA) = P
 A  (BDC ) , C   ( ADB )
Z
A
C
A
Y
X
Figure 6
 (AZB) ∩ (AYD) = AA , (CXD) ∩ (DYA)
= CC 
B
Prove:
AX BY CZ DW



1
XB YC ZD WA
Construct auxiliary segment DB (see figure 6). Next, construct the plane that
contains both AA and CC  . This plane can be constructed because AA and CC  are
coplanar, since they intersect at P. The plane will intersect DB at a point T, so it may be
referred to as “(ATC).” AT and CT will pass through  C and A , respectively, because of
the definition of (ATC).
Applying Ceva’s theorem to ADB and CDB , we get the following two equations:
(2a.)
AW DT BX


1
WD TB XA
(2b.)
BT DZ CY


1
TD ZC YB
Multiply the two equations and the result is equation (1). Q.E.D.
Now the converse must be proven.
Given:
 Space quadrilateral ABCD
X  AB , Y  BC , Z  CD , W  DA

AX
BY CZ DW




1
XB YC ZD WA
A  (BDC ) , C   ( ADB )


(AZB) ∩ (AYD) = AA , (CXD) ∩ (DYA) = CC 
Prove: (AZB) ∩ (BWC) ∩ (CXD) ∩ (DYA) = P
Construct a segment from A to a point T on auxiliary segment DB such that AT
passes through  C . Ceva’s theorem then states that equation (2a) is true. Multiply equation
(2a) with equation (1), which is given, and the result is equation (2b). Using Ceva’s theorem
once more, the result is that CT passes through A  . This result leads to the conclusion that
both AA and CC  are on the plane (ATC), because the endpoints of each segment are on the
plane. Furthermore, AA and CC  intersect at a point P because C lies in the half-plane
opposite from C  if AA is the boundary. However, because AA and CC  were defined as the
intersection lines of (ADY) and (AZB), (BWC) and (DCX), respectively, then in fact all four
planes intersect at point P. Q.E.D
Corollary of Ceva’s Theorem in Space
Theorem 7
On each face of a tetrahedron, let there be three concurrent Cevians and let every
Cevian on a face intersect a Cevian from another face on the common edge. Then the
segments connecting each vertex of the tetrahedron with the intersection point of the
Cevians on the opposite face are concurrent.
This corollary is similar to the stereometric analogy of Ceva’s theorem that was
proven above. Using the theorem, it is easy to prove this statement if one realizes that Cevians
of adjacent faces meeting on the common edge are basically intersections of planes with the
faces of the tetrahedron. Equation (1) is correct in this situation by the same logic by which it
was derived. Therefore, according to theorem 6, (AZB) ∩ (BWC) ∩ (CXD) ∩ (DYA) = P. It
must be shown that the intersection of all these planes and (DUB) ∩ (ATC) is still P (see
figure 7). Geometrically, this means that it must be proven that TU , the intersection of
(DUB) and (ATC), passes through P.
The stereometric extension of the
D
theorem of Menelaus applied to
quadrilateral ABCT as it is cut by plane
XDY states that
Z
T
AX BY CA TC 
W



 1.
XB YC AT C A
A
P
C
Divide this equation by the equation from
theorem
for
ABC ,
C Ceva’s
A
AX BY CU


 1 , and the result proves
X
Y
XB YC UA
Figure 7
the concurrency of TU , CC  , and AA at
B
P. This is equivalent to saying that all six
planes intersect at P. Therefore, the lines of intersection of the planes intersect at P.
Segments with vertices of the tetrahedron and points of intersection of the Cevians as
endpoints are parts of lines of intersection of the planes, meaning that the segments also
intersect at P. Q.E.D.
U
Applications of Ceva’s Theorem in Space and its Corollary
There are important consequences of theorems two and three. Just as Ceva’s theorem
on the plane is used to prove the concurrency of several important segments in the triangle,
Ceva’s theorem in space and its corollary can be used to prove similar things in the
tetrahedron.
Theorem 8
On each face of a tetrahedron, let the three medians be drawn. Then the segments
connecting each vertex of the tetrahedron with the intersection point of the medians on
the opposite face are concurrent.
All the conditions of theorem 7 are easily met. The medians on each face are
concurrent and the medians from adjacent faces clearly intercept the common edge at the
same point. Therefore, theorem 8 is proven by theorem 7. It is interesting to add that this
formulation is equivalent to saying that the planes drawn from each edge of the tetrahedron
and bisecting the opposite edge intersect at one point.
Theorem 9
Let a sphere intersect a tetrahedron in such a way so that it is tangent to every edge. On
every face, draw Cevians connecting to the points of tangency with the sphere on the
opposite edge. Connect the points of intersection of the Cevians on each face with the
vertex opposite the face. These new segments will intersect at one point.
It is important first of all to notice that this theorem is a three-dimensional analogue of
the Gergonne point theorem. Since the intersection of the sphere with each face is a circle
tangent to the edges, the segments connecting the vertex of the face with the point of tangency
on the opposite side of the face are concurrent. This meets the first condition of theorem 7.
Secondly, because the sphere can only be tangent to each edge at one point, Cevians to the
same edge of the tetrahedron will intersect at the same point on the edge. Therefore, both
conditions for theorem 7 are met and so the lines are concurrent.
Theorem 10
The dihedral angle bisectors of a tetrahedron meet at one point, which is the center of
the sphere inscribed in the tetrahedron.
By the stereometric angle bisector theorem, the following equations are true (figure 7
may be used):
AX S ACD
BY S ABD
CZ S ABC
DW S BCD



,
,
, and
.

YC S ACD
ZD S ABD
WA S ABC
XB S BCD
When these equations are multiplied, the result is equation (1). Hence theorem 6 states that
the four planes AZB, BWC, CXD, and DYA intersect in one point. It is now necessary to show
that planes DUB and ATC intersect at that point as well. To do that, the same reasoning as in
theorem 7 may be used. However, its assumption that the Cevians of ABC are concurrent
must first be proven:
AX BY CU S ACD S ABD S BCD





1
XB YC UA S BCD S ACD S ABD
Ceva’s theorem then states that these segments are concurrent and thus the logic in theorem 7
shows that all six dihedral angle bisectors intersect at one point. Furthermore, because the
dihedral angle bisectors are the loci of all points equidistant from two faces of a tetrahedron,
the point of intersection of the bisectors is the point equidistant from all four faces. This
means that a sphere centered at that point may be inscribed in the tetrahedron.
Theorem 11
The altitudes of a tetrahedron are concurrent if and only if the tetrahedron is
orthocentric*.
Before proceeding to prove this theorem, a lemma is necessary.
Lemma
If a tetrahedron is orthocentric then
I. A segment is an altitude of the
tetrahedron when the segment is
drawn from a vertex perpendicular
to the height on the opposite face.
II.
The heights of two faces to the same
C
W
base are concurrent.
D
A
III. The heights of a face intersect at the
Y
point where the altitude from the
B
opposite vertex intercepts the face.
A. Segment AY is perpendicular to BC by construction (See figure 8) and AD is
perpendicular to it by the definition of an orthocentric tetrahedron. Because BC is
perpendicular to two segments on (ADY), it is perpendicular to the whole plane. Since
Figure 8
*
D
In an orthocentric tetrahedron, opposite edges are perpendicular.
DD is on this plane, DD is also perpendicular to BC . By construction, DD is also
perpendicular to AY . Both of these segments are on (ABC), so DD is perpendicular to
all of (ABC), meaning it is the altitude.
B. The three perpendiculars theorem directly proves this based on the result of part A and
the definition of AY .
C. The three perpendiculars theorem directly proves this based on the result of part A and
the fact that BD is perpendicular to AC . Q.E.D.
Parts B and C of the lemma in combination with theorem 7 now easily prove the
theorem. Simply construct heights on every face of the tetrahedron and construct the altitude
from the intersection of the heights to the opposite vertex. Theorem 7 states, then, that the
altitudes intersect. Q.E.D.
Discussion
As can be seen, Ceva’s theorem in space has a great number of applications similar to
those of Ceva’s theorem on the plane. On the plane, Ceva’s theorem is used to prove that
certain interesting lines in a triangle intersect at a special point. Similarly, in a tetrahedron in
space, certain interesting planes—or segments from vertices to the opposite plane, depending
on one’s frame of reference—intersect at a special point. Hence Ceva’s theorem is useful in
exploring interesting points in the tetrahedron like the center or Gergonne point. There are
many other fascinating applications that are extensions of the ways Ceva’s theorem is used on
the plane. With more time, it would have been possible to apply it to these other interesting
three-dimensional extensions.
In our examination of the literature, we found less applications of the theorem of
Menelaus than of the Ceva’s theorem, though it does have a few interesting ones.
Unfortunately, there was not sufficient time to apply the three-dimensional extension of the
theorem of Menelaus found during research to generalized spatial problems.
Conclusions
The purpose of this project was to generalize two closely related, interesting theorems,
Ceva’s theorem and Menelaus’s theorem, to three-dimensions and then to use them in
problem solving. An extension of the theorem of Menelaus was found during examination of
the literature. The main result of the project was discovering an extension of Ceva’s theorem.
This endeavor was aided by the duality between the two theorems. The last part of the project
was to apply the newfound Ceva’s theorem. The power of this spatial extension was shown
by its simple solutions to these somewhat difficult problems. Interesting properties of the
tetrahedron were easily proven with the help of the spatial Ceva’s theorem and its corollary.
These properties included the fact that planes from each edge of the tetrahedron that bisect the
opposite edge intersect at one point, and that planes bisecting each dihedral angle intersect at
one point. Also, if a sphere is tangent to each edge of the tetrahedron, then all the planes from
each edge to the point of tangency on the opposite edge intersect at one point. Lastly, in an
orthocentric tetrahedron, the altitudes meet at one point. All of these applications are
interesting stereometric analogues in themselves of planar properties of triangles.
Acknowledgements
I would like to greatly thank my mentor for this project, Mr. Boris Koichu, for his
guidance and assistance throughout the project. His ideas were always useful. I would also
like to extend my gratitude to Ralph Tandetsky for his constant encouragement, and to the
organizers of SciTech 2001 for making this project possible.
References
1. Coxeter, H.S.M and Greitzer, S.L. Geometry Revisited, Mathematical Association of
America: 1967.
2. Lines, L. Solid Geometry, with chapters on space-lattices, sphere-packs and crystals.
Dover Publications, New York: 1965.
3. Bogomolny, Alexander. Cut the Knot! Website. <http://www.cut-theknot.com/Generalization/Menelaus.html>.
4. ---- <http://www.cut-the-knot.com/Generalization/ceva.html>.
5. O'Connor, J.J. and Robertson E.F. Menelaus of Alexandria. The MacTutor History of
Mathematics Archive. University of St. Andrews, Scotland: 2000. Website.
<http://www-history.mcs.st-and.ac.uk/~history/Mathematicians/Menelaus.html>.
6. ---- Giovanni Ceva. The MacTutor History of Mathematics Archive. University of St.
Andrews, Scotland: 2000. Website. <http://www-history.mcs.stand.ac.uk/~history/Mathematicians/Giovanni-Ceva.html>.