[proofplan] We prove holomorphicity of $f$ locally by constructing an antiderivative on each disc $B(z_0, r) \subseteq U$. The vanishing triangle-integral hypothesis and the convexity of $B(z_0, r)$ allow the [Antiderivative Existence Characterisation](/theorems/340) to produce a holomorphic $F$ with $F' = f$. Since $F$ is holomorphic, [Taylor's theorem for holomorphic functions](/theorems/348) and the [Holomorphicity of Power Series](/theorems/335) show that $F$ (and hence $f = F'$) is representable by a convergent power series, establishing holomorphicity of $f$. [/proofplan] [step:Reduce to a local problem on a disc] Fix $z_0 \in U$ and choose $r > 0$ with $B(z_0, r) \subseteq U$. The hypothesis gives $\int_{\partial T} f(z) \, dz = 0$ for every triangle $T \subseteq U$; in particular, this holds for every triangle $T \subseteq B(z_0, r)$. [/step] [step:Construct an antiderivative on $B(z_0, r)$ using the Antiderivative Existence Characterisation] The disc $B(z_0, r)$ is convex, hence star-shaped. Since $f$ is continuous on $B(z_0, r)$ and $\int_{\partial T} f(z) \, dz = 0$ for every triangle $T \subseteq B(z_0, r)$, the [Antiderivative Existence Characterisation](/theorems/340) provides a holomorphic function $F: B(z_0, r) \to \mathbb{C}$ with $F' = f$. [guided] The [Antiderivative Existence Characterisation](/theorems/340) states that a continuous function on a domain has an antiderivative if and only if its integral around every closed path vanishes. The key link is: on a convex (hence star-shaped) domain, the vanishing of triangle integrals implies the vanishing of all closed-path integrals. This is because the proof of [Cauchy's theorem for star-shaped domains](/theorems/342) constructs an antiderivative $F(z) = \int_{[z_0, z]} f(w) \, dw$ from a star centre $z_0$ along straight-line segments, using only the triangle-integral condition. Specifically, for any $z \in B(z_0, r)$ and small $h$, the triangle with vertices $z_0, z, z+h$ lies in the convex set $B(z_0, r)$. The vanishing of the triangle integral gives $F(z+h) - F(z) = \int_{[z,z+h]} f(w) \, dw$, and differentiating yields $F' = f$. The disc $B(z_0, r)$ is convex, so the star-shaped construction applies directly. Once $F$ exists with $F' = f$, the [Fundamental Theorem of Contour Integration](/theorems/339) gives $\int_\gamma f \, dz = F(\gamma(b)) - F(\gamma(a)) = 0$ for every closed path $\gamma$ in $B(z_0, r)$. [/guided] [/step] [step:Conclude that $f$ is holomorphic as the derivative of a power series] Since $F$ is holomorphic on $B(z_0, r)$, [Taylor's theorem for holomorphic functions](/theorems/348) gives $F(z) = \sum_{n=0}^\infty c_n (z - z_0)^n$ on $B(z_0, r)$. By the [Holomorphicity of Power Series](/theorems/335), the derivative $f = F'$ is also given by a convergent power series on $B(z_0, r)$: \begin{align*} f(z) = F'(z) = \sum_{n=1}^\infty n c_n (z - z_0)^{n-1}. \end{align*} In particular, $f$ is holomorphic at $z_0$. Since $z_0 \in U$ was arbitrary, $f$ is holomorphic on $U$. [/step]