Geometry question with squares

Question

Apologies for a crude drawing, but let's say we have a figure:

Where the square has the points $B(0,0)$, $A(1,0)$, $D(1,1)$, $C(0,1)$, the length of $EB$ is $\frac{1}{\sqrt{2}}$. How can we prove that if we draw a ray from $E$ parallel to $BD$ (the diagnal of the square) it will intersect $A$?

It seems intuitively obvious to me but I do not understand how to make the proof. Thank you.

EDIT: One of the given things was not actually given, I assumed it incorrectly. I apologize. I've removed it.

Answer 1

Clearly, $|EA| = \frac{1}{\sqrt{2}}$ so $\Delta AEB$ is an isosceles right angle triangle. So $\angle EAB = 45^\circ$. Since $\angle EAB = \angle ABD$, we can conclude that $AE // BD$.

Answer 2

With $EB=\frac{1}{\sqrt{2}}$, $\angle AEB=90^\circ$ and $AB=1$ given, you can apply Pythagoras' theorem to $\triangle AEB$ and calculate $AE=\sqrt{1^2-\left(\frac{1}{\sqrt{2}}\right)^2}=\frac{1}{\sqrt{2}}=EB$, so $\triangle AEB$ is isosceles, and $\angle EAB=\angle EBA=45^\circ$. Because $\angle ABD=45^\circ$ too, we can notice that $\angle EAB$ and $\angle ABD$ are 'alternate angles' and so it follows that $EA\parallel BD$.

Answer 3

Consider the two diagonals AC and BD which intersect in O, then show that $\triangle COD\cong \triangle BEA$.