Ceva's Theorem
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
Ceva's Theorem
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.)
Proof 1
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 |
= (AH·BC·AG)/(BC·AG·AH) | |
= 1 |
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
Finally
BC/D'C = BC/DC
which immediately implies D'C=DC. That is, D' and D are one and the same point.
Q.E.D.
Proof 2
Triangles CKD and BKD have a common altitude h_{K} 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.
Proof 3
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 -- A_{c}B_{c}||AB, B_{a}C_{a}||BC, and A_{b}C_{b}||AC, as shown in the diagram.
First off, since say, triangles ACC' and B_{c}CK are similar as are triangles BCC' and A_{c}CK, we have
AC'/B_{c}K = CC'/CK, and C'B/KA_{c} = CC'/CK,
which gives
(5) | AC'/C'B = B_{c}K/KA_{c}. |
From similar triangles ABB' and B_{c}KB' we get
B_{c}K/AB = KB'/BB',
while similarity of triangles ABA' and KA_{c}A' yields
KA_{c}/AB = KA'/AA'.
The latter two identities combine into
B_{c}K/KA_{c} = KB'/KA' : BB'/AA',
or, taking (5) into account,
(6_{c}) | AC'/C'B = KB'/KA' : BB'/AA'. |
Cyclically, we also have
(6_{a}) | BA'/A'C = KC'/KB' : CC'/BB' and |
(6_{b}) | 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.
Remark 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.
Remark 2
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.
Remark 3
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.
Remark 4
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.
Proof
Medians connect vertices with the midpoints of the opposite sides. Therefore,
Corollary 2 (incenter)
In a triangle, angle bisectors intersect at a single point.
Proof
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.
Proof
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 | |
= 1. |
Remark
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.)
Proof
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 AS_{a}, BS_{b}, CS_{c} intersect at a point (known as the Lemoine point.)
Proof
We'll make use of two ways to compute the area of a triangle. Namely,
Area(ΔBAS_{a})/Area(ΔAM_{a}C) = BS_{a}/CM_{a} = AB·AS_{a}/AM_{a}·AC
Area(ΔAS_{a}C)/Area(ΔAM_{a}B) = CS_{a}/BM_{a} = AC·AS_{a}/AM_{a}·AB
Divide the first of these by the second:
BS_{a}/CM_{a} · BM_{a}/CS_{a} = AB^{2}/AC^{2}
Or, since BM_{a} = CM_{a},
BS_{a}/CS_{a} = AB^{2}/AC^{2}
Similar identities hold for the other two vertices. All that remains is to multiply the three. (The symmedians have many interesting properties.)
Remark
Corollary 5 actually showed more than it set out to. The result is in fact more general. Let AP_{a}, BP_{b}, and CP_{c} be three concurrent Cevians. Reflect the line of AP_{a} in the bisector of angle A, and denote the resulting segment as AQ_{a}. Construct similarly BQ_{b} and CQ_{c} as reflections of the other two Cevians. Then the three lines AQ_{a}, BQ_{b}, and CQ_{c} are concurrent.
The lines AP_{a} and AQ_{a} 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 AP_{a}, BP_{b}, and CP_{c} and that of Cevians AQ_{a}, BQ_{b}, and CQ_{c}, are also said to be isogonal conjugates of each other.
Corollary 6
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.
Proof
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 X_{a} be the point of tangency of side BC and the excircle with center I_{a}. Similarly define points X_{b} and X_{c} on sides AC and AB. Then three lines AX_{a}, BX_{b} and CX_{c} are concurrent.
Proof
Point X_{a} 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.
9 December 2015, Created with GeoGebra
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)
Remark
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? |
Barycenter
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 w_{L} and M with mass w_{M} is point N with mass
N = Z(L, w_{L}; M, w_{M})
for the center of gravity (the barycenter) of two material points L and M with masses w_{L} and w_{M}, 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), |
(W_{1}) | There are w_{A}, w_{B} and w_{C} with pairwise barycenters at D, E, F |
(W_{K}) | There are w_{A}, w_{B} and w_{C} with the barycenter at K |
Ceva's theorem asserts that (1) and (K) are equivalent. (W_{1}) and (W_{K}) are equivalent by the assumption of the independence of the barycenter of calculations, meaning that
Z(A, w_{A}; B, w_{B}; C, w_{C}) | = Z(A, w_{A}; D, w_{B} + w_{C}) | |
(W) | = Z(B, w_{B}; E, w_{C} + w_{A}) | |
= Z(C, w_{C}; F, w_{A} + w_{B}). |
(1) implies (W_{1})
Let w_{A}> 0 be arbitrary. Find w_{B} from
w_{C}CE = w_{A}CE·AF·BD/FB/DC = w_{A}EA
by (1).
(W_{1}) implies (1)
Indeed, if D = Z(B, w_{B}; C, w_{C}) and similarly for E and F, then
BD/DC | = w_{C}/w_{B}, | |
CE/EA | = w_{A}/w_{A}, | |
AF/FB | = w_{B}/w_{C}. |
Multiplying the three gives (1).
(K) implies (W_{K})
Find w_{B} and w_{C} as to make D = Z(B, w_{B}; C, w_{C}). Find w_{A} 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), w_{A}, w_{B}, w_{C} as to make
(W_{K}) 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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