Let $ABCD$ be a trapezoid s. t. $AB\parallel DC,\ |AB|=2|DC|$ and $AD\perp BD$. Let $P$ be the midpoint of $\overline{AB},\ E, F, G$ be the intersection points of $AD\ \&\ BC,\ PC\ \&\ BD$ and $EF\ \&\ DC$ respectively.
Prove that $BG$ is passing through the midpoint of $\overline{DE}$.
My attempt:
$\overline{DC}$ is the midsegment of $\triangle ABE$.
$|AB|=2|DC|\implies |AP|=|PB|=|DC|$, so quadrilaterals $APCD, PBCD$ and $PCED$ are parallelograms. $AD\perp BD\ \land AD\parallel PC\implies\ BD\perp PC$ so, $PBCD$ is a rhombus and $\triangle APD,\triangle PBC,\triangle DCE$ and $\triangle ABE$ are isosceles.
At first glance, I thought I should expand the triangles to get another parallelogram, e.g., a parallelogram with the diagonals $\overline{DC}$ and $\overline{BH}$, where $H$ is the intersection point of $PD$ and $BG$, but that is a bit circular.
Then I thought of using the intercept theorem, but I can't apply it directly to the segment $\overline{DE}$.
Since I have to prove $BG$ bisects the segment $\overline{DE}$ I shouldn't assume a median of $\triangle BCD$ belongs to the line $BG$.
Picture:
May I ask for advice on how to solve this task?
Thank you in advance!
Hint: Show that $G$ is the centroid of $BDE$.
Note: I didn't use $BD \perp AD$, so I might have made a mistake.