# Cevian triangle of the Cevian triangle

Cevian triangle of the Cevian triangle

July 13, 2012

For a given point P in the plane of a ΔABC, the feet of the cevians through P form a ΔA'B'C' known as the cevian triangle of P with respect to the ΔABC. The reference triangle and its cevian triangle are therefore perspective with respect to the cevian point P. For example, medial ΔMaMbMc (midpoints of the sides a, b and c) is cevian triangle with respect to the centroid G.

Theorem 1: Cevian triangle of the medial triangle is perspective with the reference ΔABC.

Proof: Let Ma, Mb and Mc be midpoints of sides of ΔABC, let P be a point into the medial ΔMaMbMc, let IJK be the cevian triangle of the medial triangle with respect to point P, i.e. I = MbMc \cap MaP. Similarly for points J and K. We claim that lines AI, BJ and CK are concurrent.

Let AI\capBC = L, BJ\capAC = M, CK\capAB = N. Since IJK is cevian triangle of the medial triangle, we have (Ceva's Theorem)

Attach:Geometry/cevian1.png Δ| Cevian Δinto the medial triangle

(McI/IMb)\cdot(MbK/KMa)\cdot(MaJ/JMc) = 1

Furthermore, lines MbMc and BC are parallel, which means that McI/IMb = BL/LC (Thales). Similar, MbK/KMa = AN/NB and MaJ/JMc = CM/MA. So that,

(BL/LC)\cdot(CM/MA)\cdot(AN/NB) = 1

so three lines AI, BJ and CK intersect at a single point O. In other words, the reference ΔABC and the Cevian Δof the medial ΔIJK are perspective (in O).

A more general statement is also true.

Theorem 2: Cevian triangle of the cevian triangle is perspective with the reference ΔABC.

Proof: Let ABC is a triangle with cevian ΔEFG and Cevian point S. There is a projective transformation (a central collineation), that transforms ΔABC into ΔA'B'C' and the cevian triangle EFG into the medial ΔMaMbMc of ΔA'B'C'.

Attach:Geometry/cevian2.png Δ| A central collineation

A perspective collineation is determined by the center, axis and the vanishing line. Since the triangles ABC and EFG are perspective in a point, they are perspective in a line h\infty. Let h\infty be the line at infinity and point S be the center of the collineation. In that case, the lines EF and AB intersect on h\infty, implying that their images A'B' and MaMb are parallel. Thus the image of EFG is the medial Δof A'B'C'.

Since perspective transformations preserve incidence, Theorem 2 follows from Theorem 1.

Application 1: Medial triangle of the orthic triangle is perspective with the reference triangle. The point of perspective is the symmedian point.

Attach:Geometry/cevian3.png Δ| The symmedian point K

Application 2: The triangle whose vertices lie on the midpoints of internal sides of the parallel hexagon is perspective with reference triangle.

Proof: Let AbAcBcBaCaCb a hexagon, inscribed in ΔABC, where AbAc || CbBc , CaCb || BaAb , BaBc || CaAc. Let La , Lb , Lc midpoints of segments BaCa , CbAb , AcBc (external sides) respectively. Let Na , Nb , Nc midpoints of segments AbAc , BaBc , CaCb (internal sides) respectively.

Attach:Geometry/cevian4.png Δ| Parallel hexagon

First, we will prove that ΔLaLbLc, whose vertices lie on the midpoints of external sides of the parallel hexagon is perspective with reference ΔABC.

From Thales' theorem ALb/ALc = AbCb/AcBc, BLc/BLa = BcAc/BaCa and CLa/CLb=BaCa/AbCb.

If we multiply this, we get

(ALb/ALc)\cdot(BLc/BLa)\cdot(CLa/CLb) = 1

and after rearrangement

(ALb/LbC)\cdot(CLa/LaB)\cdot(BLc/LcA) = 1

It means (Ceva's theorem) that triangles ABC and LaLbLc are perspective in a point, implying that ΔLaLbLc is cevian triangle of the ΔABC.

According Theorem 2, the medial triangle of the ΔLaLbLc is perspective with ΔABC; but since sides of ΔLaLbLc are parallel with internal sides of the hexagon (for example, segment LbLc is the midline of the trapezoid AbAcBcCb, so LbLc || AbAc ), which proves the proposition.

Note: The configuration discussed by Vladimir is known as a Cevian Nest. For a different treatment, see a separate page.