**Cevian triangle of the Cevian triangle**

**Vladimir Nikolin****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 ΔM_{a}M_{b}M_{c} (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 M_{a}, M_{b} and M_{c} be midpoints of sides of ΔABC, let P be a point into the medial ΔM_{a}M_{b}M_{c}, let IJK be the cevian triangle of the medial triangle with respect to point P, i.e. I = M_{b}M_{c} \cap M_{a}P. 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)

_{c}I/IM

_{b})\cdot(M

_{b}K/KM

_{a})\cdot(M

_{a}J/JM

_{c}) = 1

Furthermore, lines M_{b}M_{c} and BC are parallel, which means that M_{c}I/IM_{b} = BL/LC (Thales). Similar, M_{b}K/KM_{a} = AN/NB and M_{a}J/JM_{c} = CM/MA. So that,

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 ΔM_{a}M_{b}M_{c} of ΔA'B'C'.

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 M_{a}M_{b} 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.

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

**Proof**: Let A_{b}A_{c}B_{c}B_{a}C_{a}C_{b} a hexagon, inscribed in ΔABC, where A_{b}A_{c} || C_{b}B_{c} , C_{a}C_{b} || B_{a}A_{b} , B_{a}B_{c} || C_{a}A_{c}. Let L_{a} , L_{b} , L_{c} midpoints of segments B_{a}C_{a} , C_{b}A_{b} , A_{c}B_{c} (external sides) respectively. Let N_{a} , N_{b} , N_{c} midpoints of segments A_{b}A_{c} , B_{a}B_{c} , C_{a}C_{b} (internal sides) respectively.

First, we will prove that ΔL_{a}L_{b}L_{c}, whose vertices lie on the midpoints of external sides of the parallel hexagon is perspective with reference ΔABC.

From Thales' theorem AL_{b}/AL_{c} = A_{b}C_{b}/A_{c}B_{c}, BL_{c}/BL_{a} = B_{c}A_{c}/B_{a}C_{a} and CL_{a}/CL_{b}=B_{a}C_{a}/A_{b}C_{b}.

If we multiply this, we get

_{b}/AL

_{c})\cdot(BL

_{c}/BL

_{a})\cdot(CL

_{a}/CL

_{b}) = 1

and after rearrangement

_{b}/L

_{b}C)\cdot(CL

_{a}/L

_{a}B)\cdot(BL

_{c}/L

_{c}A) = 1

It means (Ceva's theorem) that triangles ABC and L_{a}L_{b}L_{c} are perspective in a point, implying that ΔL_{a}L_{b}L_{c} is cevian triangle of the ΔABC.

According Theorem 2, the medial triangle of the ΔL_{a}L_{b}L_{c} is perspective with ΔABC; but since sides of ΔL_{a}L_{b}L_{c} are parallel with internal sides of the hexagon (for example, segment L_{b}L_{c} is the midline of the trapezoid A_{b}A_{c}B_{c}C_{b}, so L_{b}L_{c} || A_{b}A_{c} ), which proves the proposition.

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