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Ceva's TheoremGiovanni 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.
Ceva's TheoremIn a triangle ABC, three lines AD, BE and CF intersect at a single point K if and only if
(The lines that meet at a point are said to be concurrent.) Proof 1Extend 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:
from the last two we conclude that AG/BD = AH/DC and, hence,
Multiplying the identities marked with (*) we get
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
On the other hand, it's given that
Combining the two we get
Finally
which immediately implies D'C=DC. That is, D' and D are one and the same point. Q.E.D. Proof 2Triangles CKD and BKD have a common altitude hK from K. For their areas we therefore have
Similarly, on considering triangles ACD and ABD,
From (2) and (3) we derive
The latter is a key identity because two similar ones could be written starting with the other two sides:
All we need now is to multiply the three identities. Proof 3This proof is by Darij Grinberg and appeared at the geometry-college newsgroup. It is also available at his personal site.
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
which gives
From similar triangles ABB' and BcKB' we get
while similarity of triangles ABA' and KAcA' yields
The latter two identities combine into
or, taking (5) into account,
Cyclically, we also have
The product of the three is the Ceva identity
Remark 1Ceva's theorem is the reason lines in a triangle joining a vertex with a point on the opposite side are known as Cevians. Remark 2The 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 3The 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 4An 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. ProofMedians connect vertices with the midpoints of the opposite sides. Therefore, Corollary 2 (incenter)In a triangle, angle bisectors intersect at a single point. ProofFor angle bisectors we know that AF/FB = AC/BC, BD/DC = AB/AC, In a triangle, altitudes intersect at a single point. Indeed, right-angled triangles ACD and BCE are similar. Therefore
RemarkIt'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.) ProofSides 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.) ProofWe'll make use of two ways to compute the area of a triangle. Namely,
Divide the first of these by the second:
Or, since BMa = CMa,
Similar identities hold for the other two vertices. All that remains is to multiply the three. (The symmedians have many interesting properties.) RemarkCorollary 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. Corollary 6For 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. ProofIndeed, 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. ProofPoint 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.
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) RemarkCeva's theorem is implied by the theorem of Menelaus to which in fact it is equivalent. It also admits a very nice visual proof.
BarycenterThe 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 BC in the sense I plan to discuss at a later time. 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
for the center of gravity (the barycenter) of two material points L and M with masses wL and wM, respectively. Returning to our case, the center of gravity K of the two points A and D satisfies Actually this is easily done. Let wA> 0 be arbitrary. Find wB from
by (1). Now, let K be an arbitrary point inside ΔABC. Through K draw three lines, AD, BE, and CF. Ceva's theorem guarantees condition (1). From the foregoing discussion, there exist three masses wA, wB, and wC such that, if placed at the corresponding vertices of ΔABC, their center of gravity (barycenter) will coincide with the point K.  Barycenter and Barycentric Coordinates
Copyright © 1996-2008 Alexander Bogomolny |
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