How do we prove the Pizza Theorem?
I tried a coordinate bash (I also involved the concept of finding areas through definite integration)… But was too complicated.
I read about it at the following link: http://en.wikipedia.org/wiki/Pizza_theorem
Geometry – Proof of the Pizza Theorem
geometry
Related Solutions
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.
Let $A,B,C$ lie on the unit circle, with complex coordinates $a,b,c$. Since you know EGMO, you should be able to find a proof of the fact that the line through the points $a,b$ on the unit circle has equation $$z + ab\overline{z} = a+b$$
Now denoting $D = AM \cap (ABC)$ with coordinate $d$, the line $AD$ is given by $z + ad \overline{z} = a+d$.
However you know that the midpoint of $BC$, namely $ \frac{b + c}{2}$ lies on line $AD$, so
$$\frac{b + c}{2} + ad \frac{\overline{b} + \overline{c}}{2} = a+d$$
You can now just solve for $d$ to find the required intersection.
Best Answer
Lemma. Consider $n\geq2$ equally spaced chords through the point $(a,0)$ of the unit disk, having slopes $\phi_k:=\phi_0+{k\pi\over n}$ $\>(1\leq k\leq n)$. Then the sum of the squares of the $2n$ resulting chord pieces is $2n$, independently of $\phi_0$.
Proof. Intersecting the line $s\mapsto (a+s\cos\phi, s\sin\phi)$ with the unit circle we get the equation $s^2+2as\cos\phi+a^2-1=0$. If $s_1$ and $s_2$ are its two solutions we can say that $$s_1^2+s_2^2=(s_1+s_2)^2-2s_1s_2=4a^2 \cos^2\phi-2(a^2-1)=2+2a^2\cos(2\phi).$$ Noting that $\sum_{k=1}^n\cos(2\phi_k)=0$ we conclude that the sum of the $2n$ considered squares is $2n$.
Proof of the Theorem. Assume that we have $4n\geq8$ equally spaced blades, and let $S(\phi)$ denote the shaded pizza area when one blade points in direction $\phi$. When $\phi\mapsto r(\phi)$ is the polar representation of the circular pizza boundary with respect to the center of the cutter then it is easy to see that $$S'(\phi)=\pm{1\over2}\sum_{k=1}^{4n} (-1)^k r^2\left(\phi+{k\pi\over2n}\right)\ .$$ Here the right side is $\equiv0$, since the alternating sum of the $4n$ blade-length-squares vanishes according to the Lemma. It is then easy to see that $S(\phi)\equiv$ half the area of the pizza.
Applying the Theorem to two concentric pizzas of radii $1$ and $1+\epsilon$ one concludes that the "crust area" is halved as well, and in the limit $\epsilon\to0+$ it follows that the shaded areas together cover half the circumference of the pizza.