97-A generalization Zeeman-Gossard perpector

Let $ABC$ be a triangle, let a line $L$ meets the Euler line at $D$ and the line $L$ meets three sidelines $BC,CA,AB$ at $A_0,B_0,C_0$ respectively. Let $H_a, H_b,H_c$ be the orthocenter of three triangles $AB_0C_0, BC_0A_0, CA_0B_0$ respectively.  Let $G_a,G_b,G_c$ be the centroid of three triangles $AB_0C_0, BC_0A_0, CA_0B_0$ respectively. Let three points  $D_a,D_b,D_c$ lie on the Euler line of $AB_0C_0, BC_0A_0, CA_0B_0$ respectively such that:

\[ \frac{\overline{D_aH_a}}{\overline{DaGa}}=\frac{\overline{D_bH_b}}{\overline{D_bG_b}}=\frac{\overline{D_cH_c}}{\overline{D_cG_c}} \space=\frac{\overline{DH}}{\overline{DG}}=t  \]

 Let $L_a,L_b,L_c$ = three lines through $D_a,D_b,D_c$ and parallel to $BC,CA,AB$ respectively.

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Problem 1: Then triangle bounded by three line $(L_a,L_b,L_c)$ are congruent
and homothety to triangle ABC. The homothetic center lie on the line L.
When $L$ is the Euler line or t = ∞ the problem is Zeeman-Gossard perpector

Problem 2: Newton lines of four quadrilateral $(AB,AC,L_a,L)$, $(BC,BA,L_b,L)$,
$(CA,CB,L_c,L)$, $(AB,BC,CA,L)$ also through the homothetic center in problem 1

Where if $L_i$ $(i=1,2,...n)$ be a line, define $(L_1,L_2,....,L_n)$ = Polygon bound by $L_1,L_2,....,L_n$