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  "body": "\n## Proof\n\nWe identify the plane with the [[complex numbers|09FTAET5Z4]], the point $(x, y)$ with $x + iy$.\n\n1. For $u = u_1 + i u_2$ and $w = w_1 + i w_2$, with the [[complex conjugate|0FTH2E9TAS]] $\\bar u$, we have $\\bar u\\, w = (u_1 w_1 + u_2 w_2) + i(u_1 w_2 - u_2 w_1)$. This is $u \\cdot w + i \\det(u, w)$, with the [[dot product|0S52TNNTH9]] and the [[determinant|00KDZRW4C0]].\n2. Let $u, w \\neq 0$, and suppose $\\bar u\\, w = \\rho\\, e^{i\\theta}$ with a real $\\rho \u003e 0$. Taking the [[modulus|0KW2X0TFBX]] gives $\\rho = \\abs{u}\\abs{w}$, since $\\abs{e^{i\\theta}} = 1$.\n3. Dividing step 1 by $\\rho$ and using [[Euler's formula|04R3CY41VJ]] gives $\\cos\\theta = \\frac{u \\cdot w}{\\abs{u}\\abs{w}}$ and $\\sin\\theta = \\frac{\\det(u, w)}{\\abs{u}\\abs{w}}$. So $\\theta$ represents the [[oriented angle|03YN70NK0Z]] from $u$ to $w$.\n4. Let $r \u003e 0$ be the radius of the [[circle|060W5CBZP6]]. Put $a = A - O$, $b = B - O$ and $v = V - O$, each of modulus $r$.\n5. By the [[polar form|0BRXFNQ5JV]], $a = r e^{i\\alpha}$, $b = r e^{i\\beta}$ and $v = r e^{i\\nu}$ for some real $\\alpha$, $\\beta$, $\\nu$.\n6. By the [[product of exponentials|0WM7DJCM6G]], $\\bar a\\, b = r^2 e^{i(\\beta - \\alpha)}$. By step 3, $\\measuredangle AOB = \\beta - \\alpha$ modulo $2\\pi$.\n7. For real $x$ and $y$, put $m = \\frac{x + y}{2}$ and $h = \\frac{x - y}{2}$. Then $e^{ix} - e^{iy} = e^{im}(e^{ih} - e^{-ih}) = 2i \\sin(h)\\, e^{im}$ by Euler's formula.\n8. By step 7, $a - v = 2ir\\, s_a\\, e^{i(\\alpha + \\nu)/2}$ and $b - v = 2ir\\, s_b\\, e^{i(\\beta + \\nu)/2}$. Here $s_a = \\sin\\frac{\\alpha - \\nu}{2}$ and $s_b = \\sin\\frac{\\beta - \\nu}{2}$.\n9. Hence $\\overline{(a - v)}\\,(b - v) = (-2ir)(2ir)\\, s_a s_b\\, e^{i(\\beta - \\alpha)/2} = s\\, e^{i(\\beta - \\alpha)/2}$, with the real number $s = 4r^2 s_a s_b$.\n10. The vectors $a - v = A - V$ and $b - v = B - V$ are nonzero, since $V \\neq A$ and $V \\neq B$. So $s \\neq 0$.\n11. If $s \u003e 0$, step 3 with $\\rho = s$ gives $\\measuredangle AVB = \\frac{\\beta - \\alpha}{2}$ modulo $2\\pi$.\n12. If $s \u003c 0$, then $s = \\abs{s} e^{i\\pi}$, so step 9 reads $\\abs{s} e^{i((\\beta - \\alpha)/2 + \\pi)}$. Step 3 gives $\\measuredangle AVB = \\frac{\\beta - \\alpha}{2} + \\pi$ modulo $2\\pi$.\n13. Doubling a class modulo $2\\pi$ gives a well defined class modulo $2\\pi$. In both cases, $2\\,\\measuredangle AVB = \\beta - \\alpha$ modulo $2\\pi$, because $2\\pi \\equiv 0$.\n14. By step 6, $\\measuredangle AOB = 2\\,\\measuredangle AVB$.\n\n## Idea\n\nOn a circle, the chord from $v$ to $a$ points in the direction halfway between the angles of $a$ and $v$, up to a sign. So the turn from one chord to the other is half the turn from $a$ to $b$, up to a possible extra $\\pi$. Doubling removes that extra $\\pi$.\n",
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