Weaker version of Goursats theroem

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I have a homework problem that states: Let $\Omega \subseteq \mathbb{C}$ open. Suppose that $f:\Omega \rightarrow \mathbb{C}$ is holomorphic and $C^1$. Show that:

$\int_{T}f dz = 0$

Where $T$ is an arbitrary triangle contianed in $\Omega$.

My attempt so far:

$\textbf{Proof:}$

Write $f=u+iv$ and identify $\mathbb{C}$ with $\mathbb{R}^2, \ \tilde{f}:\Omega \rightarrow \mathbb{R}^2, \begin{pmatrix} x \\ y \end{pmatrix} \rightarrow \begin{pmatrix} u(x,y) \\ v(x,y) \end{pmatrix}$. Denote $\Delta$ the surface bounded by $T$ in the plane. Using Greens theorem:

$\int_{T}f dz = \int_{T} \langle \tilde{f}, \text{dx}\rangle = \int_{\Delta}\partial_x v -\partial_y u \text{dvol} = 0$, using Cauchy-Riemann.

Does this work as a valid proof?