TITLE: Dividing a disc into equal parts
QUESTION [7 upvotes]: Prove that it is not possible to divide a disc into $7$ parts of equal area by means of three straight lines.

Background: I saw this question asked in a way which seemed to imply the possibility of a simple solution.  The hint was given: what is the area on each side of a line?  Well, obviously, $\frac37$ and $\frac47$ of the whole area, but I can't see where that gets me.  Am I missing something really simple?
I tried doing some trig but it's a pain.  Even "obvious" things seem difficult to prove rigorously, for example, that the lines intersect each other at $60^\circ$ angles.
Note that the division is possible if the disc is replaced by a (suitable) different shape.  Therefore we must actually use the fact that the shape is a disc.  (Or perhaps just the fact that it is convex?)
Any ideas?  No lengthy trigonometric solutions please, I'm sure I could do that myself if I could be bothered to spend more time on it.
Please note that I am not asking for a ruler and compass construction, so this question and answer is not relevant.

REPLY [2 votes]: Each line splits the disk into parts of areas $3/7 A$ and $4/7 A$, where $A$ is the area of the disk, and that determines the distance $d$ from the centre of the disk to the line.  The general arrangement has to be something like 

where the intersection of two lines is in the $4/7 A$ region for the other line.
In order for the three wedge-shaped regions to have the same areas, the angles between each pair of lines has to be the same (so $\pi/3$).
That should be enough information to determine the area of the central triangle, say.
If it's not $1/7 A$, you're done.