Question on Theorem $V 1.1$ from Hartshone I'm about finish the prove of Theorem $V 1.1$ from Hartshorne's Algebraic Geometry, but can't really get through it:
Let $X$ a nonsingular curve, and given $C,D\in DivX$, he defines 
$C.D=C'.D'-C'.F'-E'.D+E'.F'$, 
where it have been fixed an ample divisor $H$ on X and found $n>0$ such that $nH$ is very ample divisor, then take
$C'\in|C+nH|$
$D'\in |D+nH|$ transversal to $C'$
$E'\in |nH|$ transversal to $D'$
$F'\in |nH|$ transversal to $C'$ and $E'$
It has been checked the good definition, and the properties 
$(2)$ symmetry
$(3)$ additivity
$(4)$ dependence only on the linear equivalence classes
but for $(1)$, that says: "if $C$ and $D$ are nonsingular curves meeting transversally, then $C.D=\#(C\cap D)$ the number of points of $C\cap D$" it is suggested to use the fact that if $C$ and $D$ meet transversally, then $\#(C\cap D)=deg(\mathcal{L}(D)\otimes \mathcal{O}_C)$.
I have try to use that but get stuck
$C'.D'-C'.F'-E'.D+E'.F'=\#(C'\cap D')-\#(C'\cap F')-\#(E'\cap D')+\#(E'\cap F')$
$=deg(\mathcal{L}('D)\otimes \mathcal{O}_{C'})-deg(\mathcal{L}(F')\otimes \mathcal{O}_{C'})-deg(\mathcal{L}(D')\otimes \mathcal{O}_{E'})+deg(\mathcal{L}(F')\otimes \mathcal{O}_{E'})$
then I suppose I should use additivity of degree on a curve to get to $deg(\mathcal{L}(D)\otimes \mathcal{O}_C)$, but don't really know how to do it properly.
Any suggestion is appreciated
 A: By definition, 
$$C\cdot D = C'\cdot D' - C'\cdot F' - E'\cdot D + E'\cdot F' $$
$$= \deg(\mathcal{L}(D') \otimes \mathcal{O}_{C'}) - \deg(\mathcal{L}(F') \otimes \mathcal{O}_{C'}) - \deg(\mathcal{L}(D') \otimes \mathcal{O}_{E'}) + \deg(\mathcal{L}(F') \otimes \mathcal{O}_{E'}) $$
$$= \deg(\mathcal{L}(D'-F') \otimes \mathcal{O}_{C'}) - \deg(\mathcal{L}(D'-F') \otimes \mathcal{O}_{E'})$$
By additivity of degree. 
Since $D\sim D'-F'$, you have that this is equal to 
$$ \deg(\mathcal{L}(D) \otimes \mathcal{O}_{C'}) - \deg(\mathcal{L}(D) \otimes \mathcal{O}_{E'}) $$
Now note that we may choose $C'$ and $E'$ such that $C',E'$ intersects $D$ transversally, by V.1.2 and the fact that we already know the product is well-defined, so you have that $\deg(\mathcal{L}(D) \otimes \mathcal{O}_{C'}) = \# C' \cap D = \deg (\mathcal{L}(C') \otimes \mathcal{O}_D) $ and likewise $\deg(\mathcal{L}(D) \otimes \mathcal{O}_{E'}) = \# E' \cap D = \deg (\mathcal{L}(E') \otimes \mathcal{O}_D)$. 
so this is equal to $$\deg(\mathcal{L}(C') \otimes \mathcal{O}_{D}) - \deg(\mathcal{L}(E') \otimes \mathcal{O}_{D}) $$
$$ = deg(\mathcal{L}(C'-E') \otimes \mathcal{O}_D)$$ 
$$ = deg(\mathcal{L}(C) \otimes \mathcal{O}_D)$$
