I found a conjecture generalizing the Zeeman Gossard theorem a year ago, but I haven't found a solution for this conjecture. I'm an electrical engineer, I am not a mathematician. I don't know how to prove this result. Could you give a proof?
Zeeman-Gossard theorem: Let $ABC$ be a triangle, the three Euler lines of the triangle formed by the Euler line and the sides, taken by pairs, of a given triangle, form a triangle homothetic and congruent with the given triangle and having the same Euler line.
A generalization of Zeeman-Gossard theorem: Let $ABC$ be a triangle, Let $P_1,P_2$ be two points on the plane, the line $P_1P_2$ meets $BC, CA, AB$ at $A_0,B_0,C_0$ respectively. Let $A_1$ be a point on the plane such that $B_0A_1$ parallel to $CP_1$, $C_0A_1$ parallel to $BP_1$. Define $B_1, C_1$ cyclically. Let $A_2$ be a point on the plane such that $B_0A_2$ is parallel to $CP_2$, $C_0A_2$ is parallel to $BP_2$. Define $B_2, C_2$ cyclically. The triangle formed by the three lines $A_1A_2,B_1B_2,C_1C_2$ is homothetic and congruent to $ABC$, the homothety center lies on $P_1P_2$. When $P_1P_2$ is parallel to the Euler line, this problem is the Zee-Man Gossard theorem.
Another conjecture (Concurrence of four Newton lines) (Same notations as in the cọnjecture above): the Newton lines of four quadrilaterals bounded by four lines $AB, AC, A_1A_2, P_1P_2$; four lines $BC, BA, B_1B_2, P_1P_2$; four lines $CA, $ $CB, C_1C_2, P_1P_2$; and four lines $AB, $ $BC, CA, P_1P_2$ pass through the homothety center.
1-http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
2-https://groups.yahoo.com/neo/groups/AdvancedPlaneGeometry/conversations/messages/2643