## 1. Perspective triangles

Consider triangle ABC a point D and the cevian triangle A1B1C1 of D (with vertices the traces of {DA,DB,DC} on opposite sides). Let also L be an arbitrary line in general position with respect to the side-lines of the triangle (i.e. no three lines out of the four pass through the same point). Let {A*, B*, C*} be correspondingly the intersection points of {B1C1, C1A1, A1B1} with L. Draw lines {AA*, BB*, CC*} and form through their intersections triangle A0B0C0. Then lines {AA0, BB0, C0} intersect at a common point D'.

I give an analytic proof using the projective basis (see ProjectiveBase.html ) consisting of the points {A,B,C,D}, later being the coordinator or unit point (i.e. representing the point with coordinates (1,1,1)). Assume that line L is given in this system by an equation of the form
ax + by + cz = 0.
Then the following calculations are straightforward using essentially only the vector-product as indicated in each case:
1) line(B1,C1) = (1,1,0)x(1,0,1) => x-y-z =0.
2) A* = (1, -1, -1)x(a,b,c) => (b-c, -(a+c), a+b).
3) line(C1,A1) = (1,1,0)x(0,1,1) => x-y+z =0.
4) B* = (1,-1,1)x(a,b,c) => (-(b+c), a-c, a+b).
5) line(A1,B1) = (1,0,1)x(0,1,1) => x+y-z =0.
6) C* = (1,1-1)x(a,b,c) = (b+c, -(a+c), b-a).
7) line(A,A*) = (1, 0, 0)x(b-c, -(a+c), a+b) = (0, a+b, a+c) (the tripple standing for the coefficients of the line).
8) line(B,B*) = (0, 1, 0)x(-(b+c), a-c, a+b) = (a+b, 0, b+c).
9) line(C,C*) = (0, 0, 1)x(b+c, -(a+c), b-a) = (a+c, b+c, 0).
10) C0: intersection (AA*, BB*) = (0, a+b, a+c)x(a+b, 0, b+c) ~ (b+c, a+c, -(a+b))  (~ meaning constant multiple of).
11) A0: intersection (BB*, CC*) = (a+b, 0, b+c)x(a+c, b+c, 0) ~ (-(b+c), a+c, a+b).
12) B0: intersection (CC*, AA*) = (a+c, b+c, 0)x(0, a+b, a+c) ~ (b+c, -(a+c), a+b).
13) line(C,C0) = (0,0,1)x(b+c,a+c,-(a+b)) = ( -(a+c), b+c, 0).
14) line(A,A0) = (1,0,0)x(-(b+c),a+c, a+b) = (0, -(a+b), a+c).
15) line(B,B0) = (0,1,0)x(b+c, -(a+c), a+b) = (a+b, 0, -(b+c)).
The determinant of the coefficients of the three last lines is zero since the matrix of coefficients is antisymmetric 3x3:

This implies that the three lines {AA0, BB0, CC0} are concurrent as stated.

## 2. Application with parallels

The property of the previous section has an important application concerning cevian and circumscribed triangles formed by drawing parallels to the sides of a cevian triangle.

Let A'B'C' be the cevian triangle of point D with respect to triangle ABC. Draw from the vertices of ABC parallels to the sides of the triangle to form triangle A0B0C0 (B0C0 parallel to B'C', etc.). Then triangles ABC and A0B0C0 are perspective i.e. lines {AA0, BB0, CC0} concure at a point D'.

The proof follows from the property of the previous section by applying it to the case in which the line L is the line at infinity, hence the points {A*, B*, C*} appearing there are here the "common" points of the pairs of parallels {(B'C',B0C0), (C'A',C0A0), (A'B',A0B0)}.

## 3. Synchronization of pivoting internal and external

Given two triangles ABC and A'B'C' we can in general inscribe and circumscribe in ABC triangles t1= A1B1C1 and t2=A2B2C2, both similar to the "prototype" A'B'C' so that their sides are parallel. Then rotate the system of {t1,t2} so that the triangles remain all the time inscribed/circumscribed and with sides corresondingly parallel. The result of previous section shows that when   the system of three lines {AA1,BB1,CC1} becomes concurrent then the same happens with the system of three lines {AA2,BB2,CC2}.

ProjectiveBase.html