Let $v_{1}, \ldots, v_{m}$ and $w$ be vectors in $\mathbb{R}^{n}$. Suppose that whenever $d \in \mathbb{R}^{n}$ satisfies $v_{i} \cdot d \leq 0$ for $i = 1, \ldots, m$, it also satisfies $w \cdot d \leq 0$. Prove that $w = \lambda_{1} v_1 + \cdots + \lambda_{m} v_{m}$ for some nonnegative real numbers $\{\lambda_{i}\}_{i=1}^m$.
I have formulated the given constraint as a linear problem: $\max_{v_{i} \cdot d \leq 0 \text{ for } i=1,\ldots,m} w \cdot d = 0$. The dual will then be $\min_{\sum_{j=1}^{n}w_{j}d_{i} \geq d_{i} \text{ for } i=1,\ldots,m} 0$. How should I proceed and prove the proposition (all that I can gather is that $\sum_{j=1}^{n}w_{i}\geq1$), given that the dual is correct?
If the primal LP is to maximize $\sum_{j=1}^n w_j d_j$ subject to \begin{align} \sum_{j=1}^n v_{i,j} d_j &\le 0 &&\text{for $i\in\{1,\dots,m\}$} \\ \end{align} then the dual LP is to minimize $\sum_{i=1}^m 0 \lambda_i$ subject to \begin{align} \sum_{i=1}^m v_{i,j} \lambda_i &= w_j &&\text{for $j\in\{1,\dots,n\}$} \tag1\\ \lambda_i &\ge 0 &&\text{for $i\in\{1,\dots,m\}$} \end{align}
We are given that $$\sum_{j=1}^n w_j d_j \le 0,$$ so the trivial feasible solution $d_j^*=0$ is optimal. By strong duality, the dual problem is also optimal, yielding the existence of the desired $\lambda_i$ via dual equality constraint $(1)$.