Draw two perpendicular lines, such that line1 & line2 intersect at P & line1 is at some angle theta (say 0) to x-axis.
Number the 4 quadrants Q1,Q2,Q3,Q4.
Move the intersection P along line1 , from very far left to very far right. At some position, the area on the left must equal the area on the right, because at very far left, the area of the 2 left quadrants will be 0, while at very far right, the area of the 2 right quadrants will be 0, and at some intermediate position, the areas will be equal.
Do this movement for P moving along line2 , from very far up to very far down. Again the 2 upper quadrants will be equal to the 2 lower quadrant at some point along line2.
Basically , we have moved P all over the plane , when line1 was at theta, and we have found a position where Q1+Q2=Q3+Q4=A/2 & Q2+Q3=Q4+Q1=A/2.
Now Q1=Q3 & Q2=Q4, or opposite quadrants are equal.
We get Q1=A/2-Q2.
If Q1=Q2, we are done. If not Q1 is-less-than Q2 Or Q1 is-more-than Q2.
Now repeat this process, but use theta from 0 to 90 Degrees. The locus of the points (for each theta) with this movement must be continuous.
Starting at theta=0 and going to theta=90, we see that Q1 and Q2 exchange their final values. So, starting at Q1 is-less-than Q2, we get Q1 is-more-than Q2. So, at some theta, Q1=Q2. So all 4 quadrants are equal at this point of intersection, at this angle.
Summary : We search over all intersection points, at all angles, which will give us at least one solution, because of continuity.
Here, start with the horizontal gray line and the vertical rid line on the left. Intersection is P, with 4 quadrants. We see that area is Q1+Q4=0, considering the areas of the pizza in the quadrants.
We move P slowly towards right, to get the center red line and the rightmost left line. We end up with Q2+Q3=0.
So, at some point, we have Q1+Q4=Q2+Q3, shown as the central red line.
Now, move the gray line upwards and similarly get a new P which has Q1+Q4=Q2+Q3.
In the example given, the locus of P which has Q1+Q4=Q2+Q3 is the central red line, but in general, it may be some other continuous curve C.
In the curve C, when we move P from top to bottom, we start with Q1+Q2=0 and end with Q3+Q4=0, so at some point, Q1+Q2=Q3+Q4.
If Q1=Q2 here , we are done.
Else we must repeat the process with a new theta.
When we vary theta from 0 to 90, we will see that we move from Q1 is-less-than Q2 to Q1 is-more-than Q2 , so at some intermediate point, Q1=Q2, so all 4 quadrants are equal.