Proof of Bernoulli's' result on a construction to divide any triangle into four equal parts with two perpendicular lines. I have been reading about Jacob Bernoulli and came across this particular contribution of his. Although I have tried my best to search proofs of this result I have had no success so far. Probably this is due to the fact that this theorem has no name. Any insights or sources regarding a proof or an exposition would be much appreciated. 
Thank you.
 A: Thanks for asking this question. I've had a few pleasant evenings, trying to solve this question together with a friend. We didn't actually find a construction for the triangle, but managed to prove the following more general statement:

Given any convex (compact) shape $\Delta$ in $\mathbb{R}^2$, there exists a pair of perpendicular lines dividing this shape in four parts, having the same area.

To prove this, we choose a point $P$ on the boundary of $\Delta$. We can now determine the unique point $Q$ on the boundary of $\Delta$ such that $PQ$ divides the area of $\Delta$ in two equal parts. Now, there exists a unique line $l$ that is perpendicular to $PQ$ which also divides $\Delta$ in two parts.$^*$  Let's say the intersection points of $l$ with $\Delta$ are $R$ and $S$. Since the shape is convex, the intersection of $l$ and $PQ$ lies within $\Delta$.
Seen from the point $P$, we now have four different parts of $\Delta$: counterclockwise we have southeast, northeast, northwest, southwest. We can abbreviate their areas as $SE$, $NE$, $NW$, $SW$. If you haven't made a drawing yet, I'd strongly advise you to do so now. Due to our construction, we know that $SE + NE = SW + NW$ and $SE + SW = NE + NW$.  As a result, $NE  =SW$ and $NW = SE$. We will call the area of the current southwest $A$ and the area of the currend southeast $B$. We can even define the function $\phi$ from the boundary of $\Delta$ to $\mathbb{R}$ by $\phi(P) = B - A$.
We can now move the point $P$ along the boundary of $\Delta$, possibly around corners. The whole construction of $P$, $Q$, $R$ and $S$ can still be made along the boundary of $\Delta$. We can move the point $P$ until it becomes $R$.
The construction, starting from $R$ now yields the same lines as it did, when starting from $P$. However, the values of $B$ and $A$ have switched and thus $\phi(R) = - \phi(P)$. Since the value of $\phi$ is continuous along the boundary of $\Delta$, it must have been $0$ at some point between $P$ and $R$. A this point, $SE-SW$ must have been zero and hence all four quarters must have had the same area.
$^*$ To prove this, choose a line perpendicular to $PQ$ outside of $\Delta$ and consider the difference between the area on one side of it and the other. This difference starts positive (namely, the area of $\Delta$) and ends at minus this vaule. At some point inbetween, it must have been zero. 
