Giovanni Ceva (1648-1734) proved a theorem bearing his name that is seldom mentioned in Elementary Geometry courses. It's a regrettable fact because not only it unifies several other more fortunate statements but its proof is actually as simple as that of the less general theorems. Additionally, the general approach affords, as is often the case, rich grounds for further meaningful explorations.
27 November 2015, Created with GeoGebra
In a triangle ABC, three lines AD, BE and CF intersect at a single point K if and only if
|(1)||AF/FB · BD/DC · CE/EA = 1|
(The lines that meet at a point are said to be concurrent.)
Extend the lines BE and CF beyond the triangle until they meet GH, the line through A parallel to BC. There are several pairs of similar triangles: AHF and BCF, AEG and BCE, AGK and BDK, CDK and AHK. From these and in that order we derive the following proportions:
AF/FB = AH/BC (*)
CE/EA = BC/AG (*)
AG/BD = AK/DK
AH/DC = AK/DK
from the last two we conclude that AG/BD = AH/DC and, hence,
BD/DC = AG/AH (*).
Multiplying the identities marked with (*) we get
|AF/FB · BD/DC · CE/EA||= AH/BC · BC/AG · AG/AH|
Therefore, if the lines AD, BE and CF intersect at a single point K, the identity (1) does hold. Which is to say that the fact of the three lines intersecting at one point is sufficient for the condition (1) to hold. Let us now prove that it's also necessary. This would constitute the second part of the theorem. In other words, let us prove that if (1) holds then AD, BE, CF are concurrent.
Indeed, assume that K is the point of intersection of BE and CF and draw the line AK until its intersection with BC at a point D'. Then, from the just proven part of the theorem it follows that
AF/FB · BD'/D'C · CE/EA = 1
On the other hand, it's given that
AF/FB · BD/DC · CE/EA = 1
Combining the two we get
BD'/D'C = BD/DC or
BD'/D'C + 1= BD/DC + 1 or
(BD' + D'C)/D'C = (BD + DC)/DC
BC/D'C = BC/DC
which immediately implies D'C=DC. That is, D' and D are one and the same point.
Triangles CKD and BKD have a common altitude hK from K. For their areas we therefore have
|(2)||BD/DC = Area(ΔBKD)/Area(ΔCKD)|
Similarly, on considering triangles ACD and ABD,
|(3)||BD/DC = Area(ΔABD)/Area(ΔACD)|
From (2) and (3) we derive
|(4)||BD/DC = Area(ΔAKB)/Area(ΔAKC)|
The latter is a key identity because two similar ones could be written starting with the other two sides:
AF/FB = Area(ΔAKC)/Area(ΔBKC)
CE/EA = Area(ΔBKC)/Area(ΔAKB).
All we need now is to multiply the three identities.
This proof is by Darij Grinberg and appeared at the geometry-college newsgroup. It is also available at his personal site (that since disappeared).
For this proof I changed the notations somewhat. The three lines through the point K are now AA', BB' and CC'. Draw through K three lines -- AcBc||AB, BaCa||BC, and AbCb||AC, as shown in the diagram.
First off, since say, triangles ACC' and BcCK are similar as are triangles BCC' and AcCK, we have
AC'/BcK = CC'/CK, and C'B/KAc = CC'/CK,
|(5)||AC'/C'B = BcK/KAc.|
From similar triangles ABB' and BcKB' we get
BcK/AB = KB'/BB',
while similarity of triangles ABA' and KAcA' yields
KAc/AB = KA'/AA'.
The latter two identities combine into
BcK/KAc = KB'/KA' : BB'/AA',
or, taking (5) into account,
|(6c)||AC'/C'B = KB'/KA' : BB'/AA'.|
Cyclically, we also have
|(6a)||BA'/A'C = KC'/KB' : CC'/BB' and|
|(6b)||CB'/B'A = KA'/KC' : AA'/CC'.|
The product of the three is the Ceva identity
AC'/C'B · BA'/A'C · CB'/B'A = 1.
Ceva's theorem is the reason lines in a triangle joining a vertex with a point on the opposite side are known as Cevians.
The points D, E, F may lie as well on extensions of the corresponding sides of the triangle, while the point of intersection K of the three cevians may lie outside the triangle. The proof remains the same for all possible configurations as long as all the points involved remain finite. Please look into this circumstance.
The theorem remains valid also if the lines AD, BE and CF are all parallel (in which case it's customary to say that the point K lies at infinity). This case is even simpler than the one just proven. Another exceptional case is when one (or two) of the points D, E, or F is (are) at infinity which means that one of the Cevians is parallel to the side it's supposed to cross. This case too must be treated separately.
An additional proof - a derivation from the 4 Travelers Problem - has been devised by Stuart Anderson.
Corollary 1 (center)
Medians in a triangle intersect at a single point.
Medians connect vertices with the midpoints of the opposite sides. Therefore,
Corollary 2 (incenter)
In a triangle, angle bisectors intersect at a single point.
For angle bisectors we know that AF/FB = AC/BC, BD/DC = AB/AC,
Corollary 3 (orthocenter)
In a triangle, altitudes intersect at a single point.
Indeed, right-angled triangles ACD and BCE are similar. Therefore
|AF/FB · BD/DC · CE/EA||= CE/DC · AF/EA · BD/FB|
|= BE/AD · CF/BE · AD/CF|
It's interesting to compare the direct proofs of the Corollaries with the ones we used for each case separately. Are the latter any easier?
Corollary 4 (Gergonne point)
Let D, E, F be the points where the inscribed circle touches the sides of the triangle ABC. Then the lines AD, BE and CF intersect at one point. (This is known as the Gergonne point, named after Joseph Diaz Gergonne (1771-1859). The ususal notation for the point is Ge.)
Sides of the triangle being tangent to the inscribed circle,
(An interactive illustration offers a convincing demonstration of the existence of Gergonne point and of an analogous property of excircles. The proof is virtually the same as in the case of the incircle. Curiously, the concurrency is observed also when one of the vertices of the base triangle is moved to infinity.)
Corollary 5 (Lemoine point)
Symmedians ASa, BSb, CSc intersect at a point (known as the Lemoine point.)
We'll make use of two ways to compute the area of a triangle. Namely,
Area(ΔBASa)/Area(ΔAMaC) = BSa/CMa = AB·ASa/AMa·AC
Area(ΔASaC)/Area(ΔAMaB) = CSa/BMa = AC·ASa/AMa·AB
Divide the first of these by the second:
BSa/CMa · BMa/CSa = AB2/AC2
Or, since BMa = CMa,
BSa/CSa = AB2/AC2
Similar identities hold for the other two vertices. All that remains is to multiply the three. (The symmedians have many interesting properties.)
Corollary 5 actually showed more than it set out to. The result is in fact more general. Let APa, BPb, and CPc be three concurrent Cevians. Reflect the line of APa in the bisector of angle A, and denote the resulting segment as AQa. Construct similarly BQb and CQc as reflections of the other two Cevians. Then the three lines AQa, BQb, and CQc are concurrent.
The lines APa and AQa are isogonal (or isogonal conjugates of each other.) The same is true of the other two pairs. For this reason the two points of concurrency, that of Cevians APa, BPb, and CPc and that of Cevians AQa, BQb, and CQc, are also said to be isogonal conjugates of each other.
For three concurrent Cevians AD, BE, and CF, if the points D, E, and F are reflected in the midpoints of the corresponding sides, the resulting three lines form another triplet of concurrent Cevians. In other words, isotomic conjugates of concurrent Cevians are also concurrent.
Indeed, that the given lines are concurrent is reflected by the fact (1) that the product of the three ratios is 1. Now, note that reflection in the midpoint of a side inverts the corresponding ratio. Obviously, the product of the three inverted ratios is still 1.
Corollary 7 (Nagel point)
Let Xa be the point of tangency of side BC and the excircle with center Ia. Similarly define points Xb and Xc on sides AC and AB. Then three lines AXa, BXb and CXc are concurrent.
Point Xa has a remarkable property of being midway from the vertex A. More accurately,
Corollary 8 (R. S. Hu)
Given three nonintersecting mutually external circles, connect the intersections of internal common tangents of each pair of circles with the center of the other circle. Then the resulting three line segments are concurrent.
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Circles B and C are homothetic in D - the point of intersection of their common internal tangents. Which means that
(See R. B. Nelsen, Proofs Without Words II, MAA, 2000)
Ceva's theorem is implied by the theorem of Menelaus to which in fact it is equivalent. It also admits a very nice visual proof.
|What if applet does not run?|
The point where three medians of a triangle meet is called the barycenter - the center of gravity - of the triangle. There is a way to justify this designation. Place equal masses w at the vertices of the triangle. Then the midpoint D of BC will be the center of gravity of the vertices B and C. Intuitively, if we place the sum of two masses in their center of gravity then the moment (the mass times the distance) of such a material point relative to any other point will equal the sum of moments of the two original points.
Generally speaking, the center of gravity of two points L with mass wL and M with mass wM is point N with mass
N = Z(L, wL; M, wM)
for the center of gravity (the barycenter) of two material points L and M with masses wL and wM, respectively. (A similar notation applies to the barycenter of any number of points.)
Returning to our case, the center of gravity K of the two points A and D satisfies
We wish to demonstrate that the following conditions are equivalent
|(1)||AF/FB · BD/DC · CE/EA = 1,|
|(K)||AD, BE, CF are concurrent (in, say, K),|
|(W1)||There are wA, wB and wC with pairwise barycenters at D, E, F|
|(WK)||There are wA, wB and wC with the barycenter at K|
Ceva's theorem asserts that (1) and (K) are equivalent. (W1) and (WK) are equivalent by the assumption of the independence of the barycenter of calculations, meaning that
|Z(A, wA; B, wB; C, wC)||= Z(A, wA; D, wB + wC)|
|(W)||= Z(B, wB; E, wC + wA)|
|= Z(C, wC; F, wA + wB).|
(1) implies (W1)
Let wA> 0 be arbitrary. Find wB from
wCCE = wACE·AF·BD/FB/DC = wAEA
(W1) implies (1)
Indeed, if D = Z(B, wB; C, wC) and similarly for E and F, then
Multiplying the three gives (1).
(K) implies (WK)
Find wB and wC as to make D = Z(B, wB; C, wC). Find wA as to make
Ram Tobolski elucidates: assume (K) holds so that the three segments AD, BE, and CF are concurrent in K. Find (as above), wA, wB, wC as to make
(WK) implies (K)
This is a direct consequence of (W). Indeed, by (W), K must lie on each of AD, BE, CF forcing them to be concurrent at K.
As we see, Ceva's theorem (the equivalence of (1) and (K)) is a quintessential expression of the existence of the barycenter of three material points, or (W), without which the barycenter would not be well defined. On the other hand, (W) is a consequence of the linearity of the definitions and can be easily proved in analytic geometry.
Barycenter and Barycentric Coordinates
- 3D Quadrilateral - a Coffin Problem
- Barycentric Coordinates
- Barycentric Coordinates: a Tool
- Barycentric Coordinates and Geometric Probability
- Ceva's Theorem
- Determinants, Area, and Barycentric Coordinates
- Maxwell Theorem via the Center of Gravity
- Bimedians in a Quadrilateral
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Three glasses puzzle
- Van Obel Theorem and Barycentric Coordinates
- 1961 IMO, Problem 4. An exercise in barycentric coordinates
- Centroids in Polygon
- Center of Gravity and Motion of Material Points
- Isotomic Reciprocity
- An Affine Property of Barycenter
- Problem in Direct Similarity
- Circles in Barycentric Coordinates
- Barycenter of Cevian Triangle
- Concurrent Chords in a Circle, Equally Inclined
Menelaus and Ceva
- The Menelaus Theorem
- Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
- Ceva's Theorem
- Ceva in Circumscribed Quadrilateral
- Ceva's Theorem: A Matter of Appreciation
- Ceva and Menelaus Meet on the Roads
- Menelaus From Ceva
- Menelaus and Ceva Theorems
- Ceva and Menelaus Theorems for Angle Bisectors
- Ceva's Theorem: Proof Without Words
- Cevian Cradle
- Cevian Cradle II
- Cevian Nest
- Cevian Triangle
- An Application of Ceva's Theorem
- Trigonometric Form of Ceva's Theorem
- Two Proofs of Menelaus Theorem
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Menelaus from 3D
- Terquem's Theorem
- Cross Points in a Polygon
- Two Cevians and Proportions in a Triangle, II
- Concurrence Not from School Geometry
- Two Triangles Inscribed in a Conic - with Elementary Solution
- From One Collinearity to Another
- Concurrence in Right Triangle
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