Proof from "An isoperimetric inequality with applications to curve shortening" by Gage Proposition 3 in the paper An isoperimetric inequality with applications to curve shortening by Michael E. Gage states the following:
For any closed, convex $C^1$ curve it is possible to choose an origin so that inequality $\int_{\gamma} p^2 ds\leq {LA\over\pi}$ holds. If the curve is piecewise $C^2$ then inequality $\pi{L\over A}\leq\int_{\gamma} \kappa^2 ds$ holds as well.
Here $p(s)=<X(s),-N(s)>$, being $X(s)$ the curve with arclength parameter $s$ and $N(s)$ the inward normal vector, and $L, A, \kappa$ the legnth of the curve, the area it encloses and its curvature respectively.
The first part of his proof is this:

For each point $X(s)$ on the boundary curve $\gamma$, there is a unique point $Y(s)$ such that the line from $X(s)$ to $Y(s)$ bisects the convex lamina bounded by $\gamma$. Define a function $f(X(s))=<(T_{X(s)}\times T_{Y(s)}),n>$ where $T_{X(s)}$ and $T_{Y(s)}$ are the tangents at $X(s)$ and $Y(s)$ and $n$ is the positively oriented normal to the plane. Since $\gamma$ is $C^1$, $f$ is continuous and in addition $f(X(s))=-f(Y(s))$. From the intermediate value theorem it follows that $f(X(s))=0$ for some $s_1$ and that $T_{X(s_1)}=-T_{Y(s)}$. Choose the origin to be the midpoint of the line from $X(s_1)$ to $Y(s_1)$ and let the $x$ axis lie along this chord.

I'm trying to understand this part involving how to correctly choose an origin:


*

*First, I don't see why it is obvious that there is a unique point $Y(s)$ such that the line from $X(s)$ to $Y(s)$ bisects the region bounded by $\gamma$. Maybe because I don't see clearly what he means when he uses the term bisect.

*Second, I'm struggling to understand what is the behaviour of the function $f$, and this is a key part of the proof, as you can see.


It would be great if anyone could explain to me a little bit these two ideas. Thanks in advance.
 A: The term 'bisects' very likely means that the region bounded by the convex curve is divided into two regions of equal area by such a line. The existence is (e.g.) a consequence of the mean value theorem applied to a family of lines passing through $X(s)$ and $X(s+t)$ for $t$ in the range $(0,\ell)$, $l$ being the length of the curve. You kind of swipe out the area of the region bounded by the curve using a family of lines through one single point. Since that region is convex such a line divides the region bounded by the curve into two subregions with continuously varying area (continuity is what you have to verify and what allows you to apply the mean value theorem). 
It is not clear to me what exactly you are asking about the behaviour of $f$, maybe you can make that more precise. 
(Edit) Note: Gage extends the geometric picture to Euclidean three space. There, the cross product of two vectors is normal to these vectors and its length is the area of the parallelogram spanned by the two vectors. This (the area of said parallelogram) is what he gets by multiplying with the normal. Note that this function measures, in some sense, how close the curve is to a circle. On a circle it vanishes identically, and (without checking) I'm sure the converse is true as well. On an ellipse which is not a circle it vanishes only in four points
I looked into the papers of Gage about some 15 years ago (I do assume that's what you are looking at), it's interesting to see this is still an area of interest.
