Trilinears            f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a [cos²(B - C)][b cos(C - A) + c cos (A - B)]
Barycentrics    af(a,b,c) : bf(b,c,a) : cf(c,a,b)

Let A'B'C' be the cevian triangle of a point X = x : y : z (trilinears), let L(A) be the line through A parallel to B'C', and define L(B) and L(C) cyclically. The lines L(A), L(B), L(C) form a triangle homothetic to A'B'C', with homothetic center

ax²(by + cz) : by²(cz + ax) : cz²(ax + by).

We denote this point by D(X) and call it the Danneels point of X. (Eric Danneels, Hyacinthos, Jan. 28, 2005).

The formula is simpler in barycentrics: if U = u : v : w, then

D(U) = u²(v + w) : v²(w + u) : w²(u + v).

If X is line the Euler line, then D(X) is on the Euler line. A proof follows. Let a₁, b₁, c₁ be cos A, cos B, cos C, respectively, so that the Euler line is given parametrically as x(t) : y(t) :z (t) by

      x = b₁c₁ + ta₁
      y = c₁a₁ + tb₁
      z = a₁b₁ + tc₁,

where t is an arbitrary function homogenous of degree 0 in a, b, c. Then D(X) is the point U = u : v : w given by

      u = ax²(by + cz)
      v = by²(cz + ax)
      w = cz²(ax + by).

A point P = p : q : r is on the Euler line if

a(b² - c²)(b² + c² - a²)p
+ b(c² - a²)(c² + a² - b²)q
+ c(a² - b²)(a² + b² -c²)r = 0.

It is easy to check that the point U satisfies this equation. (This sort of algebraic proof is more inclusive than a geometric proof, because here, a,b,c are indeterminates or variables. They can, for example, take values that are not sidelengths of a triangle. Here, cos A is defined as (b² + c² - a²)/(2bc), so that no dependence on a geometric angle is necessary.)

The appearance of (I, J) in the following list means that D(X(I)) = X(J):
(1,42)    (2,2)    (3,418)    (4,25)    (6,3051)    (7,57)    (8,200)    (69,394)    (75,321)    (100,55)    (110,84)    (264,324)    (366,367)    (651,222)    (653,196)    (1113,25)    (1114,25)    (1370,455)

X(3078) lies on this line: (2,3)

X(3078) = X(5)-Ceva conjugate of X(233)