Prove that a quadrilateral, and the quadrilateral formed by the orthocenters of four related triangles, have the same area. In a forum the following theorem was proposed, but I couldn't find a synthetic solution.

Given a quadrilateral $ABCD$, let $P$, $Q$, $R$ and $S$ be resp. the
  orthocenters of triangles $ABC$, $BCD$, $CDA$ and $DAB$. Prove that
  quadrilaterals $ABCD$ and $PQRS$ have the same area.

I found online a proof for the case $ABCD$ is cyclic, but have no clue for the general case.
EDIT.
Note that if $ABCD$ is cyclic a much stronger result holds: $ABCD$ and $PQRS$ are equal. This is problem 2.35 in Mathematical Olympiad Treasures by Andreescu and Enescu. They prove the theorem using vectors and derive it from a lemma which may be useful: if $H$ is the orthocenter of a triangle $ABC$ and $O$ its circumcenter, then one has: $\vec{OA}+\vec{OB}+\vec{OC}=\vec{OH}$.
EDIT 2.
Finding an expression for the coordinates of the orthocenter, as a function of the coordinates of the vertices, is not difficult. For instance, we get for $P$:
$$
x_P=\frac{\left(y_A-y_B\right) \left(y_C-y_A\right) \left(y_B-y_C\right)-x_A x_B
   \left(y_A-y_B\right)-x_A x_C \left(y_C-y_A\right)-x_B x_C \left(y_B-y_C\right)}{x_C
   \left(y_A-y_B\right)+x_A \left(y_B-y_C\right)+x_B \left(y_C-y_A\right)},\\
\ \\
y_P=\frac{\left(x_A-x_B\right) \left(x_C-x_A\right) \left(x_B-x_C\right)-y_A y_B
   \left(x_A-x_B\right)-y_A y_C \left(x_C-x_A\right)-y_B y_C \left(x_B-x_C\right)}{y_A
   \left(x_B-x_C\right)+y_B \left(x_C-x_A\right)+y_C \left(x_A-x_B\right)}.
$$
The signed area $S_{ABC}$ of triangle $ABC$ can be conveniently expressed as:
$$
2S_{ABC}=A\times B+B\times C+C\times A=
x_A y_B-x_B y_A +x_B y_C-x_C y_B +x_C y_A -x_A y_C,
$$
where I used the shortcut $A\times B=\det(A,B)$. Of course, the sign of $S_{ABC}$ depends on the order of the vertices.
Also notice that denominators in the above formulas for $x_P$ and $y_P$ are just $-2S_{ABC}$.
We can then define a signed area for quadrilateral $ABCD$:
$$
2S_{ABCD}=S_{ABC}+S_{BCD}+S_{CDA}+S_{DAB}.
$$
I used Mathematica to find general expressions for the signed areas of $ABCD$ and $PQRS$ as a function of the coordinates of $A$, $B$, $C$, $D$, and checked they are indeed equal. This counts as a proof, but the resulting expressions are too long and involved. Moreover, that gives no insight for the reason of that equality.
In doing the above, I also realized that a stronger equality holds, of which I hadn't been aware before. It turns out that 

The area of triangle $PQR$ is the same as the area of triangle $DAB$
  (that is of the triangle having $S$ as orthocenter), and so on.

Of course this implies that $ABCD$ and $PQRS$ have the same area. Hopefully, this stronger result could be easier to prove without using algebra and coordinates.
 A: 
To prevent this good question, which has been widely viewed but not yet answered, from expiring for lack of oxygen, I'll submit this imperfect proof. Perhaps someone can help me fill the gap, or see a way around it. Approaching the problem in its edited form, given quadrilateral $ABCD$, with diagonals $AC, BD$, and orthocenters $E,F,G$ of triangles $ABC, BCD$, and $CDA$, respectively, to prove without coordinate geometry that$$\triangle ABD=\triangle EFG$$Through $D$ and $E$ draw $DS$ and $ET$ parallel to $AB$ and $GF$, respectively, intersecting in $M$ (if they are parallel, $ABCD$ is a trapezoid and the proof is easier). Join $MA,MB,MG,MF$, and through $M$ draw $PMN$ perpendicular to $AB$, and $QMR$ perpendicular to $GF$, and join $QT$ and $PS$. Since $\angle SMP$ and $\angle TMQ$ are right angles, and $\angle PSM=\angle QTM$ (assuming that $S,T,U,V$ are concyclic), which seems to be the case but I lack proof), then triangles $MPS$ and $MQT$ are similar, making$$\frac{MT}{MS}=\frac{MQ}{MP}=\frac{MN}{MR}$$But since $AG$ and $BF$ are both perpendicular to $DC$, they are parallel, making$$\frac{MT}{MS}=\frac{GF}{AB}$$Therefore$$\frac{MN}{MR}=\frac{GF}{AB}$$and triangles $ABM$ and $MFG$, since their bases are in inverse ratio to their altitudes, are equal. But $$\triangle ABM=\triangle ABD$$since they are on the same base and in the same parallels.  For the same reason$$\triangle MFG=\triangle EFG$$Therefore$$\triangle ABD=\triangle EFG$$Note: It might seem more direct to join $SB$ and $TF$, making $\triangle ABS=\triangle ABD$ and $\triangle GFT=\triangle GFE$, and then show that triangles $ABS$ and $GFT$, between parallels $AG$ and $BF$, are on bases $AS=GT$. But however much Geogebra indicates this equality of bases, I've yet to find proof of it.
