How to divide a pizza between friends equally without using centre Here's a really fun question a friend told me abut. He claims to know the correct answer, and told me the answer, but left proving the answer as an exercise to me. Now, It's been ages since he asked me the question, and he himself seems to have forgotten the solution, so here's the question for the community(I tried searching, but useless):

Two friends go to a pizza shop and purchase one. They decide to make 8 slices of the pizza. Then the waiter comes and challenges them to cut the pizza such that the point where all cutting lines meet is NOT the centre of the pizza, but another point. The friends are jolts of each other and think that one of them will get a bigger area f pizza is this method is used. To satisfy both of them, the waiter himself cuts it in such a way that both get 4 slices, and there cumulative area is equal. How did the waiter do this ?

So, basically, you have to draw chords across the circle such that all of them intersect at a certain point(thats not the center) and then choose 4 pieces equal in area to the other 4.
My friend claims that if we divide the circle as in the image below:

IN this way, or basically any way, the following property holds:

ar(1) + ar(3) + ar(5) + ar(7) = ar(2) + ar(4) + ar(6) + ar(8)

Hence, the two friends can distribute the pizza amongst themselves.
But now, he doesn't remember how to prove this mathematically. He does say that the proof is very easy and was childish and obvious once you've read it, but he is unable to trace the book this was part of now...
So, I turn to you, friends, help me to solve this.
Thanks in advance,
Nib
 A: I don't believe the claim. Look at the figure below, in which each "thin" area is smaller than its two neighbors. Clearly the sum of the thin areas is smaller than the sum of the fat areas. 

It's possible that there's some arrangement of areas (i.e., I get two large and two small, and you get two large and two small) in which the pizza area comes out equal...but it's not "alternating odd and even". 
And if you move the "cut point" even closer to the edge, it's possible to get a situation in which one slice is more than half of the pizza, in which case there cannot possibly be a fair division. 
I think that your friend must have forgotten something. (And as Claude's comment shows...the missing thing is "the angles at the slice-center must all be equal", which the ones in your diagram are not.)
Here's a possible solution: pick a point $C$ to be the center, and let $P$ be the point of the circle near $C$. Draw the line $PC$, which will be a diameter, and then draw three more lines. Give one person all the slices on one side of $PC$, and the other one all the slices on the other side. They then each get half the pizza. This answers the original question, although not your friend's claim. A perhaps cleaner way:
Draw any diameter of the circle as your first line, and then draw three more lines that meet at some off-center point of that diameter. Then give each person the slices on one side of the diameter. 
A: One cut is a diameter.
[.. The other 3 cuts can be anything as long as they intersect on that diameter at a point other than the center of the circle (to meet the requirements of the question). Give one friend 4 slices above the diameter and the other 4 slices below. Each 4 slice group obviously will be half of the pizza regardless of how those other cuts come out.]
A: Symmetric construction wrt diameter. PizzaCut of same triangle area between same eccentric base same height between parallel lines. Add and remove, equaling sectors at Pizza circle center.

