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  "body": "\n## Proof\n\n1. Let $U \\subseteq \\C$ be open and let $f : U \\to \\C$ be continuous. Assume $\\oint_{\\partial T} f(z) \\, dz = 0$ for each closed triangle $T \\subseteq U$.\n2. For $a, b \\in \\C$, let $[a, b]$ be the segment path $t \\mapsto a + t(b - a)$, $t \\in [0, 1]$. All integrals below are [[contour integrals|00FWNP9TXY]].\n3. Reversing a segment changes the sign: $\\int_{[b, a]} f = -\\int_{[a, b]} f$. This follows from the [[substitution|0D37QFPEZE]] $t \\mapsto 1 - t$.\n4. Fix $c \\in U$ and $R \u003e 0$ with $B = \\set{z : \\abs{z - c} \u003c R} \\subseteq U$. It suffices to show that $f$ is [[holomorphic|0DG0Z72CQD]] on each such disc $B$.\n5. For $z \\in B$, define $F(z) = \\int_{[c, z]} f(w) \\, dw$. The segment $[c, z]$ lies in $B$ because $B$ is convex.\n6. Let $z \\in B$ and let $h \\ne 0$ with $z + h \\in B$. The triangle $T$ with vertices $c$, $z$, $z + h$ lies in $B$, by convexity.\n7. Claim: $\\int_{[c, z]} f + \\int_{[z, z + h]} f + \\int_{[z + h, c]} f = 0$. If the vertices are not collinear, the left side is $\\pm \\oint_{\\partial T} f$, so it is $0$ by step 1.\n8. If the vertices are collinear, they lie on a line $s \\mapsto c + s u$ with $\\abs{u} = 1$. By substitution, each segment integral becomes $\\int f(c + s u) \\, u \\, ds$ over an interval of $s$. The three intervals add to zero by additivity of the integral, which proves the claim.\n9. By steps 3 and 7, $F(z + h) - F(z) = \\int_{[z, z + h]} f(w) \\, dw$.\n10. Also $\\int_{[z, z + h]} f(z) \\, dw = \\int_0^1 f(z) \\, h \\, dt = f(z) \\, h$. So $$\\frac{F(z + h) - F(z)}{h} - f(z) = \\frac{1}{h} \\int_{[z, z + h]} \\bigl( f(w) - f(z) \\bigr) \\, dw.$$\n11. Let $\\varepsilon \u003e 0$. Since $f$ is continuous at $z$, there is $\\delta \u003e 0$ with $\\abs{f(w) - f(z)} \u003c \\varepsilon$ for $\\abs{w - z} \u003c \\delta$.\n12. For $0 \u003c \\abs{h} \u003c \\delta$, each $w$ on $[z, z + h]$ satisfies $\\abs{w - z} \u003c \\delta$. The segment has length $\\abs{h}$.\n13. By the [[estimation lemma|0DJMGPN8K9]] and step 12, the right side of step 10 has modulus at most $\\frac{1}{\\abs{h}} \\cdot \\varepsilon \\abs{h} = \\varepsilon$.\n14. So $F$ is [[complex differentiable|0D1C9Y1EXA]] at $z$ with $F'(z) = f(z)$. Thus $F$ is holomorphic on $B$ and $F' = f$ there.\n15. By [[holomorphic functions are infinitely differentiable|0QEBN8DGVE]], $F'$ is holomorphic on $B$. So $f$ is holomorphic on $B$.\n16. Each point of $U$ lies in such a disc $B$. So $f$ is complex differentiable at each point of $U$, and $f$ is holomorphic on $U$.\n\n## Idea\n\nOn a disc, integrating $f$ along segments from the center gives a function $F$. The triangle condition makes the difference $F(z + h) - F(z)$ an integral over the short segment from $z$ to $z + h$, so continuity of $f$ gives $F' = f$. The derivative of a holomorphic function is holomorphic, so $f$ is holomorphic.\n",
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