# Show that a convex compact set in $R^2$ can be cut into 4 sets of equal area by 2 perpendicular lines

Okay I need to show this using calculus and mean value theorem.

My try :

Let $D$ be a convex and compact set in $R^2$ Now let $R$ be a compact closed rectangle such that $D \subset R$. Draw two lines parallel to the axis such that $D$ is now composed of four subsets name them $R_1 \cup R_2 \cup R_3 \cup R_4 = D$

Okay now create these functions :

$$F(p)= 1 \ if\ p\epsilon R_1 \\ =0 \ else\\\\G(p)= 1 \ if\ p\epsilon R_2 \\ =0 \ else \\\\ H(p)= 1 \ if\ p\epsilon R_3 \\ =0 \ else \\\\ T(p)= 1 \ if\ p\epsilon R_4 \\ =0 \ else$$

Then we have $$\int\int_{R}F(p)+\int\int_{R}G(p)+\int\int_{R}H(p)+\int\int_{R}T(p)=A(D)$$

$A(D)$ denotes the area of this compact convex set $D$

Now since $$\int\int_{R}F(p)= A(R_1)=\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij})$$

We can say that $\exists R_{ij}: \bigcup R_{ij}=R_1$ such that $$\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij})> \frac{A(D)}{4}$$$Also$\exists R_{ij}: \bigcup R_{ij}=R_1$such that $$\sum_{R_{ij}: \bigcup R_{ij}=R_1} A(R_{ij}) < \frac{A(D)}{4}$$$