De 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.

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 hK from K. For their areas we therefore haveArea(ΔCKD) = DC·hK/2 and Area(ΔBKD) = BD·hK/2, from which
(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 .

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,
which gives
(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.
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,AF/FB = BD/DC = CE/EA = 1. Each of the ratios is 1 and so is their product.
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, CE/EA = BC/AB. Multiplying the three yields (1).
Corollary 3 (orthocenter)
In a triangle, altitudes intersect at a single point.
Proof
Indeed, right-angled triangles ACD and BCE are similar. Therefore CE/DC = BE/AD. In an analogous manner, AF/EA = CF/BE and BD/FB = AD/CF. Now
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, AF = EA, FB = BD, DC = CE so that (1) indeed holds.(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.)
Proof
We'll make use of two ways to compute the area of a triangle. Namely, 2·S = a·b·sin(C) and2·S = c·hc (and similarly for the other two vertices.) Thus we have
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.)
Remark
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.
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 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.
Proof
Point Xa has a remarkable property of being midway from the vertex A. More accurately,AB + BXa = AC + CXa. Let p be the semiperimeter of ΔABC. Then BXa = p - AB = p - c andCXa = p - AC = p - b. Therefore, BXa/CXa = (p - c)/(p - b). Write the corresponding ratios for side AB and AC and multiply all three equalities to prove the Corollary.
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.