Question about the Ham sandwich theorem. 
The question I’m referring to is question 29. As it states I have to find an prove that for any angle theta between 0 and pi , it is possible to cut the slice in half with a cut of incline theta. 
I am so confused by this question and honestly have no idea where to start. Can someone help me please?
 A: Here is how we reason when the line is not vertical: 
Let $ 0\leq \theta \leq \pi$, $\theta \neq \frac{\pi}{2}$ denote your favorite angle. Then the equation
$$y= tan(\theta)x+b$$
defines a straight line with incline $\theta$, crossing the y axis in the point $(0,b)$. If you vary the value of $b$, you will see that the line is moved vertically up and down.(but changing $b$ does not change the incline! In other words, the steepness of the line stays the same)
Now, this line divides the plane into two pieces. Call one of the pieces $P_1$ and the other $P_2$.
Now we imagine that we have a piece of ham with area $A$ lying on the plane, and define $$A(b)=\overset{\text{area of ham that lies}}{\text{in plane $P_1$}}-\overset{\text{area of ham that lies}}{\text{in plane $P_2$}}$$
Now, the function $A$ is continous(you don't need to prove this). Also, $A(b_1)=A$ and $A(b_2)=-A$ for some values $b_1$ and $b_2$ of $b$ (because we can move the line either completely to one side of the ham, or completely to the other side of the ham.)
By the IVM=intermediate Value Theorem, there is a value of $b$, say $b_0$, that is between $b_1$ and $b_2$ and such that 
$$A(b_0)=0$$
But this simply means that the line with incline $\theta$ that crosses the y -axis in the point $(0,b_0)$ divides the ham into two pieces of equal area. Since the angle $\theta$ was not specified, it is true for every $\theta.$ In other words: No matter which incline we choose, we can find a line with that incline that divides the ham in two pieces. 
