Note: I changed your notations to more sensible ones.
Suppose $x^{(j)}, j \in \{1,N\}$ are such extreme points at which optimal value of the objective function $f$ is attained. Denote such optimal value $M = f(x^{(j)})$. Let $c = (c_1,\dots,c_n)^T$. Then $f(x) = \sum_{i=1}^n c_ix_i = c^T x$. Now let $x = \sum_{j = 1}^N \lambda_j x^{(j)}$ be a convex combination of $\{x^{(1)},\dots,x^{(N)}\}$ with $\lambda_j \ge 0$ and $\sum_{j=1}^N \lambda_j = 1$.
\begin{align}
f(x) &= f\left( \sum_{j=1}^N \lambda_j x^{(j)} \right) \\
&= c^T \left( \sum_{j=1}^N \lambda_j x^{(j)} \right) \\
&= \sum_{j=1}^N \lambda_j c^T x^{(j)} \tag{$v \mapsto c^Tv$ is linear} \\
&= \sum_{j=1}^N \lambda_j f\left( x^{(j)} \right) \\
&= \sum_{j=1}^N \lambda_j M \\
&= M \tag{$\sum_{j=1}^N \lambda_j = 1$}
\end{align}
This confirms the first statement in the question body: any convex combination of "optimizers" is also an "optimizer".
Next, to address OP's doubt.
However, a theorem of extreme point is that "not a convex combination of other points". This makes me confused. If a convex combination [of two or more points in the feasible region] cannot be an extreme point then how can I get the optimal value of the objective function?
Optimal points and extreme points are two concepts. The former can be either finitely are infinitely many, but the later is finitely many (as long as the dimension of the feasible region is finite). From the Fundamental Theorem of Linear Programming, an LPP has optimal feasible solution if and only if it has optimal solution at the extreme points in the feasible region. Since we only have finitely many such extreme points, we can use an algorithm (notably the simplex algorithm) to find the optimal value of the objective function.