How to show directly that finitely generated cone contains polyhedral cone of nonnegative vectors? Let $P=\{x \in \mathbb{R}^n \mid Ax \geq b, x \geq 0 \}$ be a nonempty polyhedron for matrix $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$.
According to Minkowski-Weyl theorem $P$ can be written as 
$$
P=\text{conv}(v_1,\cdots,v_p)+ \text{cone}(d_1,\cdots,d_l)
$$
for some $v_i \in \mathbb{R}^n$ and $d_j \in \mathbb{R}^n$.
Let $C=\{x \in \mathbb{R}^n \mid Ax \geq 0, x \geq 0 \}$.
Show that $C \subseteq \text{cone}(d_1,\cdots,d_l)$.
The reverse has been proven in How to show polyhedral cone of nonnegative vectors contains finitely generated cone?
My try:
$C$ can be written as the following:
$$
Ax \geq 0, x \geq 0 \,\,\, \Rightarrow 
A' = 
\begin{bmatrix}
A
\\
I
\end{bmatrix}
x
\geq 0
$$
Hence, $C=\{x' \in \mathbb{R}^{2n} \mid A'x' \geq 0 \}$ which is a polyhedral cone and using the theorem saying that every polyhedral cone is a finitely generated cone we are done.
I want to show this using rigorous prove, and not using that theorem.

We know that 
$$
P=\text{conv}(v_1,\cdots,v_p)+ \text{cone}(d_1,\cdots,d_l)=V+D.
$$
Take $x\in C$ $\Leftrightarrow$ $Ax\ge 0$, $x\ge 0$, and $p\in P$ then
$$
A(p+tx)=Ap+tAx\ge b,\quad p+tx\ge 0,\quad\forall t\ge 0,
$$
hence $p+tx\in P$, $\forall t\ge 0$. From the representation above
$$
p+tx=v_t+d_t\quad\Leftrightarrow\quad \frac{1}{t}p+x=\frac{1}{t}v_t+\frac{1}{t}d_t.
$$
Let $t\to+\infty$ then $\frac{p}{t}\to 0$ and $\frac{v_t}{t}\to 0$ ($V$ is bounded), hence
$$
\frac{1}{t}d_t\to x.
$$
Since $\frac{d_t}{t}\in D$ and $D$ is closed, we must have $x\in D$.
