The grey area is equal to the white area Problem. Show that the sum of the areas of the white regions is equal to the sum of the areas of the grey regions. All the angles between consecutive chords are $45^\circ$.
A solution (not totally Euclidean) can be obtained as follows: Say that two perpendicularly intersected chords are split in pieces of sizes $a,b$ and $c,d$, resp. Then, it is not hard to show that
$$
a^2+b^2+c^2+d^2=4R^2
$$
where $R$ is the radius of the circle. (Proposition 11 from Archimedes' book of Lemmas.) Suppose now that we rotate all the four chords by angle $\vartheta$, then it can be readily shown that
$$
Grey(\vartheta+\Delta\vartheta)=Grey(\vartheta)+{\mathcal O}(\Delta\vartheta^2),
$$
and hence $Grey'(\vartheta)=0$.
Can we produce a solution which is suitable for High-Schools?
EDIT. The new figure corresponds to the Proof without words by L. Carter & S. Wagon (1994a), "Proof without Words: Fair Allocation of a Pizza", Mathematics Magazine 67 (4): 267.


 A: This is an attempt to give some preliminary lemmas, in order to propose to your students a minor variation of your wonderful proof. The parametric equation of a translated circle is given by:
$$(a+\cos\varphi,b+\sin\varphi) \tag{0}$$
hence the polar equation is given by:
$$ \rho(\theta)^2 = a^2+b^2+1+2a\cos\varphi+2b\sin\varphi,\qquad \tan\theta=\frac{b+\sin\varphi}{a+\cos\varphi} \tag{1}$$
and since the area in polar coordinates is simply given by $\frac{1}{2}\int \rho(\theta)^2\,d\theta$ (that is the hardest part to prove, but I believe that Cavalieri's principle and triangulations should do the job nicely), the pizza theorem boils down to showing that
$$\large\scriptstyle \left(\int_{0}^{\pi/4}+\int_{\pi/2}^{3\pi/4}+\int_{\pi}^{5\pi/4}+\int_{3\pi/2}^{7\pi/4}\right)\, \rho^2(\theta)\,d\theta = \left(\int_{\pi/4}^{\pi/2}+\int_{3\pi/4}^{\pi/2}+\int_{5\pi/4}^{3\pi/2}+\int_{7\pi/4}^{2\pi}\right)\, \rho^2(\theta)\,d\theta \tag{2}$$
Now proposition $11$ in Archimedes' book of lemmas can be read as:
$$ \forall\theta,\quad \sum_{k=0}^{3}\rho^2\!\!\left(\theta+\frac{k\pi}{2}\right)=4, \tag{3}$$
hence both the RHS and the LHS of $(2)$ equal $\color{red}{\pi}$.
In principle, we may just use $(1)$ to prove that, by naming $g(a,b)$ the $LHS$ of $(2)$,
$$\nabla g = 0 \tag{4}$$
holds, but that involves some nasty changes of variables and differentiation under the integral sign, and it is probably not suited for high school students.
