Area always half, geometry puzzle.

A cute fact about triangles with integer co-ordinates:

Lemma: Suppose $A, B, C$ are points in the 2D plane each with integer co-ordinates, such that no point with integer co-ordinates lies inside, or on the sides (except $A, B, C$) of triangle $ABC$. Then, the area of triangle $ABC$ is exactly $\dfrac{1}{2}$ 

[A point with integer co-ordinates is a point whose x and y co-ordinates are both integers].

For example, $A = (0,0)$, $B = (1,1)$, $C = (2,1)$. The "base" $BC$ and height $AD$ (where $D = (0,1)$) are both of length $1$ and so the area of $\triangle{ABC}$ is exactly $\dfrac{1}{2}$

Can you prove the Lemma?

[Note, using a stronger theorem about lattice points and areas would be circular (unless you prove the stronger theorem without using this Lemma).]

[A neat proof of this appears in Proofs from the Book: Solution]