jmc

Let the points at which the segments hit the triangle be called $D,D^{\prime },E,E^{\prime },F,F^{\prime }$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ($△ABC\sim △DP{D}^{\prime }\sim △PE{E}^{\prime }\sim △F^{\prime }PF$). The remaining three sections are parallelograms. By similar triangles, $B{E}^{\prime }=\frac{d}{510}\cdot 450=\frac{15}{17}d$ and $EC=\frac{d}{425}\cdot 450=\frac{18}{17}d$. Since $F{D}^{\prime }=BC-E{E}^{\prime }$, we have $900-\frac{33}{17}d=d$, so $d=\boxed{306}$.