Why does Euclid need the point D in the proof of proposition 3 of the book 1? 
I'm reading Euclids Elements Book 1 and get stuck at the proof of proposition 3. There he wants to prove that one can cut of a line $C$ of a line $AB$ assuming that $C$ is smaller then $AB$.

In the proof he gave us an arbitrary line $AB$ and a shorter one $C$. Then he could place the straight line $C$ at the point $A$ and got the line $AD$ (this can be done using proposition 2 of book 1). Then he draw the circle $DEF$ with center $A$ and radius $AD$. Now using the definition of a circle $AD=AE$ and since $C=AD$ he got that $AE=C$.

Now my question is, why does we need the point $D$. Wouldn't it be enough to copy the line $C$ with the compass and then get to the point E?
Thanks for your help.
 A: Euclid does allow you to copy the line $C$ to have one endpoint at $A$.
What Euclid does not allow is to use your eyeballs to very, very carefully copy the line segment $C$ in a manner which lines up exactly along the line through $A$ and $B$.
The best you can be guaranteed to do is to copy the line segment $C$ to some line segment with one endpoint at $A$, and that's the line $AD$ in the diagram. There are no guarantees regarding the position of the point $D$. You may not guarantee that it lies on the line segment $AB$.
What you then do with that line segment $AD$ is to use it to construct a circle centered on $A$ with radius $AD$, and then you intersect circle with the line that extends $AB$.
A: An addition to @LeeMosher 's excellent answer, responding to

Wouldn't it be enough to copy the line  with the compass and then
get to the point E?

That would require a  compass that retains the separation between its legs when you lift it from the plane to draw a segment elsewhere. Euclid's axioms allow only a collapsing compass. Propositions I.2 and I.3 assert that you can in fact suppose you have the rigid kind.
That's the compass equivalence theorem.
