Numerical range of a matrix contains the convex hull of the eigenvalues. I am stuck with the following question.

Question: Let $A \in \mathbb{C}^{m \times m}$ be arbitrary. Let $W(A)$ be the numerical range i.e. the set of all Rayleigh quotients of $A$ corresponding to a all nonzero vectors $x \in \mathbb{C}^m$. Show that $W(A)$ contains the convex hull of the eigenvalues of $A$.

Progress: We can write $W(A) = \{ x^TAx : \| x \| = 1 \}$. If $(\lambda_i, q_i)$ is an eigenpair of $A$, we know that the Rayleigh quotient is $r(q_i) = \frac{q_i^TAq_i}{q_i^Tq_i} = \lambda_i$. It is enough to show that every convex combination $\alpha_1\lambda_1 + \alpha_2\lambda_2 \in W(A)$ (i.e. $\alpha_1 + \alpha_2 = 1$).

Because $A \in \mathbb{C}^{m \times m}$ we don't have an orthogonal basis of eigenvectors. Since we can, I thought it would be helpfull to choose $\| q_i \| = 1$. I tried taking $x = \sqrt{\alpha_1}q_1 + \sqrt{\alpha_2}q_2$, then
$$
r(x) = \frac{\alpha_1\lambda_1 + \alpha_2\lambda_2 + \sqrt{\alpha_1\alpha_2}(\lambda_1 + \lambda_2)q_1^Tq_2}{\| \text{norm} \|^2},
$$
but then the problem with the $q$'s not being orthogonal arises.

Can somebody help me solve this problem?
The numerical range $W(A)$ is convex. This is a non-trivial fact, as far as I know, so maybe you are supposed to use this fact and not prove it.
According to Wikipedia page for Rayleigh quotient, Rayleigh quotient is usually defined for Hermitian Matrix, so you can assume that eigenvectors are orthonormal and hence the last part of your equation is equal to zero.