Simultaneous Generalization of the Theorems of Ceva and Menelaus

Ceva's and Menelaus' theorems are useful tools in establishing concurrency of lines and collinearity of points. Ceva's theorem is an easy consequence of Menelaus' and the opposite is also true. Although similar, the two theorems work in complementary circumstances. Let A1A2A3 be a triangle with points B1, B2, B3 in sidelines A2A3, A1A3, and A1A2, respectively. Menelaus' theorem holds when an even number (0 or 2) of points B are internal to the sides of ΔA1A2A3; Ceva's theorem holds otherwise, when an odd number (1 or 3) of points B are internal to the sides of the triangle.

According to Ceva's theorem, the cevians AiBi are concurrent provided

A1B3

B3A2
·
A2B1

B1A3
·
A3B2

B2A1
 = 1

According to Menelaus' theorem, the three points Bi are collinear provided

A1B3

B3A2
·
A2B1

B1A3
·
A3B2

B2A1
 = -1

The converse theorems are also true and are easily shown by contradiction to be equivalent to the direct statements. Note that all the segments involved are thought to be directed so that, for example, A2B1 = - B1A2.

Assume b1, b2, b3 are real numbers such that A2B1 = A2A3 / (1 + b1) so that B1A3 = b1A2A3 / (1 + b1), and b2 and b3 are defined similarly. For convenience, we allow b1 = -1 in case B1 is a point at infinity. In these notations,

A1B3

B3A2
·
A2B1

B1A3
·
A3B2

B2A1
 = 
1

b1b2b3

so that b1b2b3 = 1 for Ceva's and b1b2b3 = -1 for Menelaus' theorems.

We now place additional three points C1, C2, C3 on the side lines A2A3, A1A3, and A1A2. The real numbers c1, c2, c3 play the role similar to that of b1, b2, b3, with a change of direction, for example, C1A3 = A2A3 / (1 + c1), etc. Then the following theorem holds.

Theorem

C1B2, C2B3 and C3B1 are concurrent iff

(*) b1b2b3 + c1c2c3 + b1c1 + b2c2 + b3c3 = 1.

Proof

For the proof we'll use the barycentric coordinates. For any point, P in the plane of ΔA1A2A3 the barycentric coordinates p1:p2:p3 could be found starting with the signed areas of triangles A1A2P, A2A3P and A3A1P:

p1 : p2 : p3 = [A1A2P] : [A2A3P] : [A3A1P].

where, say, the area [A1A2P] is positive if triangles A1A2P and A1A2A3 have the same orientation and is negative otherwise. In other words, [A1A2P] is positive iff P is on the same side of A1A2 as A3. The barycentric coordinates are homogeneous, meaning that, for any real k≠0,

p1 : p2 : p3 = kp1 : kp2 : kp3.

In barycentric coordinates, B1(0:b1:1), B2(1:0:b2), B3(b3:1:0) and, correspondingly, C1(0:c1:1), C2(1:0:c2), C3(c3:1:0). Let X(x1:x2:x3) be a generic variable point in the plane of ΔA1A2A3. Then the lines B1C3, B2C1, and B3C2 are given by the determinant equations

 
x1 x2 x3
0 b1 1
1 c3 0
 = 0, 
x1 x2 x3
1 0 b2
0 1 c1
 = 0, 
x1 x2 x3
b3 1 0
c2 0 1
 = 0. 

These reduce, respectively, to

-c3x1 + x2 - b1x3 = 0,
-b2x1 - c1x2 + x3 = 0,
x1 - b3x2 - c2x3 = 0.

These three lines are concurrent iff their equations are linearly dependent which is only true when

 
c3 -1 b1
b2 c1 -1
-1 b3 c2
 = 0, 

which is exactly b1b2b3 + c1c2c3 + b1c1 + b2c2 + b3c3 = 1.

Observe that if C1, C2, C3 coincide with A2, A3, A1, respectively, then c1 = c2 = c3 = 0 and (*) becomes b1b2b3 = 1. This is Ceva's theorem. On the other hand, if C1, C2, C3 coincide with B1, B2, B3, respectively, then b1c1 = b2c2 = b3c3 = 1, so that (*) reduces to b1b2b3 + 1/b1b2b3 = -2 or b1b2b3 = -1. This is Menelaus' theorem.

I placed an interactive illustration of the theorems of Ceva and Menelaus on a separate page.

References

  1. M. S. Klamkin, A. Liu, Simultaneous Generalization of the Theorems of Ceva and Menelaus, Mathematics Magazine, Vol 65, No 1 (February 1992), pp. 48-52

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Menelaus and Ceva

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