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  "body": "\n## Proof\n\nWe show that the exterior angle and the interior angle at $C$ add up to $\\pi$, and then use the angle sum.\n\n1. Since $C$ lies strictly between $B$ and $D$, there is a real $\\lambda$ with $0 \u003c \\lambda \u003c 1$ and $C = (1 - \\lambda) B + \\lambda D$.\n2. Then $C - B = \\lambda (D - B)$ and $D - C = (1 - \\lambda)(D - B)$. So $D - C = -\\mu (B - C)$ with $\\mu = (1 - \\lambda)/\\lambda \u003e 0$.\n3. The vertices of the [[triangle|0CTNFGYMM2]] are distinct, so $A \\neq C$ and $B \\neq C$. By step 2, also $D \\neq C$. So the [[angles|0GJAJC6TNZ]] $\\angle ACD$ and $\\angle ACB$ are defined.\n4. By the definition of the angle, with the [[dot product|0S52TNNTH9]] and the [[length|0XKBKGTCPC]],\n$$\\cos \\angle ACD = \\frac{(A - C) \\cdot (D - C)}{\\norm{A - C} \\, \\norm{D - C}} = \\frac{-\\mu \\, (A - C) \\cdot (B - C)}{\\mu \\, \\norm{A - C} \\, \\norm{B - C}} = -\\cos \\angle ACB.$$\n5. By the [[addition formula|028XEQCE88]], [[$\\cos \\pi = -1$ and $\\sin \\pi = 0$|01A8JGR4JM]], and the evenness of the [[cosine|08WFZP6P89]], $\\cos(\\pi - \\angle ACB) = -\\cos \\angle ACB$.\n6. Both $\\angle ACD$ and $\\pi - \\angle ACB$ lie in $[0, \\pi]$, and by steps 4 and 5 they have the same cosine.\n7. The cosine is injective on $[0, \\pi]$, by the uniqueness in the definition of the [[arccosine|0CWDVN132S]]. So $\\angle ACD = \\pi - \\angle ACB$.\n8. By the [[angle sum of a triangle|05MDKQBEAF]], $\\pi - \\angle ACB = \\angle BAC + \\angle CBA$.\n9. Steps 7 and 8 give $\\angle ACD = \\angle BAC + \\angle CBA$.\n\n## Idea\n\nThe rays from $C$ to $B$ and from $C$ to $D$ point in opposite directions. So the exterior angle and the interior angle at $C$ fill a straight angle $\\pi$. The other two interior angles fill the same gap, by the angle sum.\n",
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