- Consider S, a closed convex region in R^2. Let us first target bisecting the area.
A line can be found in any direction n, which will bisect the region I.
This is because:
Consider a line which is placed outside S, then this line cuts the region I in a fraction 0:1. Now traverse this line continuosly, and parallel to itself, to other side of S. Now the line cuts the region I in ratio 1:0. Since this is a continuous traversion, there exists a position where the line cuts region in ratio 1:2.
Now, this argument is true for any given direction n, and for any convex region.
Thus there exists a line in every direction which will bisect a convex region I.
Now, for the length of S.
Consider line l, which cuts the region I into half has a direction n as it's normal. Let a and b be the lengths of the curves which are towards n and -n respectively. And let d = a-b (difference).
Now, we can turn the line continuously and get various n, by maintaining the condition that the line bisects the region I(from above). Thus when normal direction of the line is -n, then d = -(a-b).
Here, either (a-b) or -(a-b) is negative and other is positive. Thus, d has continuosly varied from positive to negative, and thus should have acquired 0(zero) at some position of n. Which states that a = b, and thus the lengths are equal.
Thus, we can find a line which will bisect area and perimeter of a convex region simultaneously.