= Solution
Let $T$ be a closed triangle contained, together with a neighbourhood, in the domain of a <holomorphic function> $f$. We prove <Cauchy theorem for a triangle> without assuming continuity of $f'$.
Join the edge midpoints to subdivide $T$ into four triangles, each with half its diameter and perimeter. Their positively oriented <contour integrals> sum to the integral around $T$, because the internal edges cancel. At least one subtriangle has integral of modulus at least one quarter of the original. Repeat that choice to obtain nested closed triangles $T_n$ satisfying
$$
\left|\int_{\partial T}f(z)\,dz\right|\leq4^n\left|\int_{\partial T_n}f(z)\,dz\right|,
\quad d_n=2^{-n}d_0,\quad P_n=2^{-n}P_0.
$$
Their intersection is a single point $z_*$, by compactness and shrinking diameter. <complex differentiability at a point> at $z_*$ gives
$$
f(z)=f(z_*)+f'(z_*)(z-z_*)+(z-z_*)r(z),\qquad r(z)\to0.
$$
The first two terms have polynomial primitives, hence integrate to zero around each triangle. For sufficiently large $n$, $|r|<\varepsilon$ on $T_n$, so the <complex line integral estimate> gives
$$
\left|\int_{\partial T_n}f(z)\,dz\right|\leq\varepsilon d_nP_n.
$$
Combining the bounds gives $|\int_{\partial T}f(z)\,dz|\leq\varepsilon d_0P_0$. Since $\varepsilon$ is arbitrary,
$$
\boxed{\int_{\partial T}f(z)\,dz=0.}
$$
The orientation can be reversed with no change to the conclusion.
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