Prove $c\vec{v}+d\vec{w}$ with $0 ≤ c,d ≤ 1$ fill the parallelogram with sides $\vec{v}$ and $\vec{w}$ 
For vectors  $\vec{v},\vec{w}\in\mathbb{R}^2$, prove that the combinations $c\vec{v}+d\vec{w}$ with $0 ≤ c ≤ 1$ and $0 ≤ d ≤ 1$ fill the parallelogram with sides $\vec{v}$ and $\vec{w}$

If I take a simple examples like $\vec{v}=(1,0)$ and $\vec{w}=(0,1)$
$$
c\begin{bmatrix}
1\\0
\end{bmatrix}+d\begin{bmatrix}
0\\1
\end{bmatrix}=\begin{bmatrix}
c\\d
\end{bmatrix}=\begin{bmatrix}
x\in[0,1]\\y\in[0,1]
\end{bmatrix}
$$
But how do I prove it mathematically for the general case ?
This is stated by @Jack D'Aurizio in the post Decide if a point is inside parallelogram, but no proof is given.
My Attempt

$$
\vec{r}=c\vec{v}+d\vec{w}\\
c,d\in[0,1]\implies c+d\in[0,2]\\
\text{when }c+d\in[0,1]\implies c\leq 1-d\\
\vec{r}\leq (1-d)\vec{v}+d\vec{w}=\vec{v}+d(\vec{w}-\vec{v}) \\
\implies\vec{r} \text{ fills the triangle }\Delta OAB\\
$$
When $c+d\in[1,2]$ how do I prove that $\vec{r}$ span the other half of the parallelogram ?
 A: 
By the figure above, $$\vec u = \vec v+ \alpha \vec w$$ with $\alpha \in [0,1]$. See that the vector $\vec u$ takes the "border" of the parallelogram. Furthermore, if we take $\beta \vec u$ (green vector) with $\beta \in [0,1]$ we get interior points of the parallelogram.
In this case, the vector $$\beta \vec u = \beta \vec v + \beta \alpha \vec w$$ when we vary $\alpha$ and $\beta$, the vector $\beta \vec u$ takes all interior points of the upper triangle shaped by $\vec v$ and $\alpha \vec w$. You just need to make the necessary changes to get the lower triangle.
Can you finish?
A: To solve by triangles, rotate by $180^\circ$ about either $\frac{\vec v+\vec w}{2}$ or the origin (i.e. consider the parallelogram $(-\vec v, -\vec w)$), and the transforms $c\to 1-c, d\to 1-d$.
$\vec r = (\vec v+\vec w) - c\vec v - d\vec w = (1-c)\vec v+(1-d)\vec w$
To continue your method, try:
$$
\vec{r}=c\vec{v}+d\vec{w}\\
c,d\in[0,1]\implies c+d\in[0,2]\\
\text{when }c+d\in[1,2]\implies 1-d\leq c\leq 1\leq 2-d\\
(1-d)\vec{v}+d\vec{w}\leq\vec{r}\leq \vec{v}+d\vec{w}\\
\text{when }c+d\in[1,2]\implies 1-c\leq d\leq 1\leq 2-c\\
c\vec{v}+(1-c)\vec{w}\leq\vec{r}\leq c\vec{v}+\vec{w}\\
\implies\vec{r} \text{ fills the triangle }\triangle ABD\\
$$
I think you need to add lower bounds to your original argument for completeness.
