What is the probability that EA > EC?

I found this question in one of my books:

Square $ABCD$ has a length of $1$. Point $E$ is selected inside the square. Segment $EA$ represents the distance between point $E$ and one of the vertices of the square, $A$, and segment $EC$ represents the distance between point $E$ one of the vertices of the square, $C$. What is the probability that $EA > EC$?

Since $E$ can be anywhere on the square, the first thing to focus on is point $E$ being on the perpendicular bisector of $AC, BD$.

If it is on $BD$, then it's equidistant from $A$ and $C$. In this case, neither $EA$ nor $EC$ are greater than each other, but it still adds a possibility to where $E$ can be.

If $E$ is anywhere away from $BD$ and closer to $A$, then $EC > EA.$

If $E$ is anywhere away from $BD$ and closer to $C$, then $EA > EC.$

To me, it seems like there are $3$ total possibilities and there's only one where $EA > EC,$ making it $1/3$ for the answer to the question. In my book, it says the possibility $EA > EC$ is $1/2$, which I don't understand because it indicates that there are $2$ total possibilities, but $E$ being on the perpendicular bisector is still a possibility.

Any help is appreciated!

You can assume that $A=(0,0), C=(1,1), E=(x,y), 0\le x,y\le 1$. Then your condition means $x^2+y^2>(x-1)^2+(y-1)^2$ or $2x+2y>2$ or $y>1-x$. So the probability is the area of the region of the square above the line $y=1-x$ which (the line) is the diagonal $BD$. That area is $1/2$, so the probability is $1/2$.

One can argue also that the probability that $AE>EC$ must be the same as the probability that $EC>AE$ because of the symmetry of the square and the sum of these probabilities is $1$ (because the probability that $EC=AE$ is $0$; it is the area of the diagonal $BD$). If you want to know more about geometric probability, look here.