The Thomson cubic is one of the famous cubic curves associated with the given triangle. It was intensively studied for many years and the evidence of these studies are the properties of Thomson cubic which can be found in (Kimberling, 1998), (Cundy, Parry, 1995), (Rubio, 1989). Most of the properties of the Thomson cubic (including its definition) have been stated in terms of locus. For example (Gibert, 2012) the Thomson cubic is the locus of points P such that trilinear polar (Coxeter, 1993) and the polar line (in the cicumcircle) (Wells, 1991) are parallel. Definition of the Thomson cubic can be stated as following. Let G be the centroid of the given triangle ABC. Then the Thomson cubic is the locus of points P such that P, P* (isogonal conjugate of P) (Sigur, 2005) and G are collinear. Using homogeneous barycentric coordinates on the plane of ΔABC (Yiu, 2 2 2 2000) with ΔABC as a reference triangle and taking into consideration that G = (1,1,1), P = (x,y,z), P* = (a²/x, b²/y, c²/z,) (a = BC, b = CA, c = AB are the lengths of sides of ΔABC) one can write the equation of the Thomson cubic as

or (Gibert, 2012)

(1)

In this article, they introduce the concept of quasi-symmedian point of the ΔABC which generalizes the concept of symmedian point and prove that the locus of quasi-symmedian points is the Thomson cubic.

Consider the point P in the plane of ΔABC and denote A’, B’ and C’ the feet of cevians AP, BP, and CP, respectively . Construct the points B₁ and C₂ such that A is the midpoint of segments BB₁ and CC₂. Draw the lines l_A, l_B, l_C through A, B and C,  respectively, which are parallel to BC, CA, and AB. Draw the lines B₁B’ and C₂C’ and denote X₁ = B₁B’ ∩ l_C and X₂ = C₂C’ ∩ l_B. Construct the pairs of points Y₁, Y₂ and Z₁, Z₂ in a similar way. Then the lines AX₁, BY₁ and CZ₁ are concurrent and so are AX₂, BY₂ and CZ₂ (Figure 1).

Figure 1. The concurrent Line of AX₁, BY₁, CZ₁ is AX₂, BY₂, CZ₂.

The authors use barycentric coordinates in a plane of ΔABC. Denote (x : y : z) for coordinates of the point and [x : y : z] of coordinates of the line. Recall that the coordinates of the line through M = (x₁ : y₁ : z₁) and M = (x₂ : y₂ : z₂) are cofactors of the matrix

and the coordinates of the common point of the lines [x₁ : y₁ : z₁] and [x₂ : y₂ : z₂] are cofactors of the matrix

Let P = (x : y : z). Then A’ = (0 : y : z), B’ = (x : 0 : z), and C’ = (x : y : 0). As it is easy to check, B₁ = (2 : -1 : 0) and C₂ = (2 : 0 : -1); l_B = [1 : 0 : 1] and l_C = [1 : 1 : 0]. Find the equation of B₁B’ =  = [z : 2z : -x] and C₂C’ =  = [y : -x : 2y]. Next find  = (-x : x : z) and  = (-x : y : -x). Using cyclic permutations, Y₁ = (x : -y : y), Y₂ = (y : -y : z), Z₁ = (z : y : -z), Z₂ = (x : z : -z). Finally,

and, in a similar way, BY₁ = [y : 0 : -x] and CZ₁ = [-y : z : 0]. The lines AX₁, BY₁, and CZ₁ are concurrent because

and the point of their concurrency is

Analogously, the lines AX₂, BY₂, and CZ₂ are concurrent at

The proof of the lemma has been completed.

Construction which has been described in the Lemma maps the point P = (x : y : z) onto the set {W₁, W₂} of two points    and . If P is the symmedian point (Kimberling, 1994) P = K = (a² : b² : c²) of ΔABC, the points  and  coincide with the first and the second Brocard points of ΔABC, Ω₁ and Ω₂ (Honsberger, 1995).

The authors assume in the previous conclusion of Lemma and the following consideration that the vertices A, B, and C of ΔABC are ordered in counterclockwise direction. Call this direction positive. Call the angle MNQ in the plane of ΔABC positive if the shortest rotation from the ray MN to the ray NQ is positive. In that sense the interior angles of ΔABC have positive measures. Denote these measures A, B, C, respectively.

By definition of Brocard points,

For the points W₁ and W₂ defined above introduce the following angles

Introduce also two numbers: Δ₁ = cot α₁ + cot β₁+ cot γ₁ and Δ₂ = cot α₂ + cot β₂ + cot γ₂ and call Δ = Δ₁ - Δ₂ the symmedian defect of the point P.

Definition :- The point P in the plane of ΔABC is its quasi-symmedian point, if the symmedian defect of P is zero: Δ = 0.

Obviously, the symmedian point of ΔABC is its quasi-symmedian one.

The main result of this article can be stated now as following.

Theorem :- The locus of quasi-symmedian points of ΔABC is its Thomson cubic.

Proof :- The authors shall prove that, for the point P, Δ₁ - Δ₂ = 0 if and only if the triple (x : y : z) satisfies the equation (1). So, let us calculate cot α_k, cot β_k, cot γ_k, k = 1,2.

Let Ax₁ ∩ BC = A’’ = (0 : xy : yz) = (0 : x : z).

• If A’’ is between B and C, then, by definition of barycentric coordinates BA’’/A’’C = z/x and it follows that BA’’ = az/z+x. On the other hand, by the law of sines, BA’’/sin α₁ = c/sin(B+ α₁) (Figure 2).
  Therefore,
  
  The last equality can be written as
  
  Or
  
  (2)

  
  where S is the area of ΔABC.
• Let A’’ be on the ray BC outside the segment BC (Figure 3). They may assume in this case that x < 0, z > 0, |x| < z. Then we have again BA’’/A’’C = z/ |x| and calculate BA’’ = az/z + x. Because ∠BAW₁ = α₁ is the measure of the interior angle of ΔABA’’ we may continue the calculations using the law of sines to get again the formula (2) for cot α₁.
• In the case presented in the Figure 4 they have ∠BAW₁ = α₁ < 0. Therefore, the exterior angle of ΔABC at A has the  positive measure -α₁, ∠W₁AB = -α₁. It follows that -α₁ = ∠B + ∠BA''A such that ∠BA''A = -B -α₁ and ∠BA''A = 180^o + α₁.  Now applying the law of sines to ΔABA''
  
  Or
  
  So, the formula (2) works in all three cases.
• There are several degenerated cases where the authors cannot apply the law of sines (or an equivalent formula for area of a triangle) because there is no triangle to be used. They can apply barycentric coordinates of points which are points of a projective plane, not Euclidian plane. Degenerated cases can appear when some of such points as P, B’, W₁, or A’’ are at infinity. Applying the idea of continuity of baricentric coordinates and the value of α₁ from a position of the point P (if P is not 1 the vertex of ΔABC), they can conclude that the formula (2) is true in such cases, too.

Figure 2. The Sines Law.

Figure 3. Outside Segment of BC ray.

Figure 4. The exterior ∠BAW

Using (2) and cycling permutations cot β₁ = a² (x+y)/2Sx - cot C and γ₁ = b² (y+z)/2Sy - cot A. That gives

Analogously, cot α₂ = b² (x+y)/2Sy - cot C,

Hence,

It follows that Δ₁ - Δ₂ = 0 if and only if

Note that the last equation was obtained assuming that each coordinate of the point P = (x : y : z) is different from zero. But if, for example, x = 0, then it follows from (1) that yz(z - y) = 0, so y = 0 or z = 0 or y = z. That gives the vertices and the midpoints of the sides of ΔABC, and one may consider these points as additional quasi-symmedian points for which P lies on lines AB, BC, CA.