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  "body": "\n## Proof\n\nWe reduce both directions to one identity for dot products with unit vectors.\n\n1. Let $a = d(B, C)$, $b = d(A, C)$ and $c = d(A, B)$, with the [[Euclidean distance|0NPPGHQNXQ]] $d$. The vertices of the [[triangle|0CTNFGYMM2]] are distinct, so $a, b, c \u003e 0$.\n2. Since $D$ is a point of the [[segment|VQ4Q2V45RP]] $BC$, there is $t \\in [0, 1]$ with $D = (1 - t) B + t C$.\n3. Then $D - B = t (C - B)$ and $D - C = (1 - t)(B - C)$. So $d(B, D) = t a$ and $d(D, C) = (1 - t) a$.\n4. We have $D \\neq A$. Otherwise $A$ lies on the line $BC$, and the vertices are [[collinear|0AFNNTRTJE]]. So the [[angles|0GJAJC6TNZ]] $\\angle BAD$ and $\\angle DAC$ are defined.\n5. Let $u = B - A$, $v = C - A$ and $w = D - A = (1 - t) u + t v$. Let $e = u / c$ and $f = v / b$. These have [[length|0XKBKGTCPC]] $1$.\n6. By the definition of the angle, with the [[dot product|0S52TNNTH9]], $\\cos \\angle BAD = \\dfrac{u \\cdot w}{c \\, \\norm{w}} = \\dfrac{e \\cdot w}{\\norm{w}}$ and $\\cos \\angle DAC = \\dfrac{w \\cdot v}{\\norm{w} \\, b} = \\dfrac{f \\cdot w}{\\norm{w}}$.\n7. Both angles lie in $[0, \\pi]$, where the cosine is injective by the definition of the [[arccosine|0CWDVN132S]]. So $\\angle BAD = \\angle DAC$ if and only if $(e - f) \\cdot w = 0$.\n8. By step 5, $w = (1 - t) c \\, e + t b \\, f$. Since $e \\cdot e = f \\cdot f = 1$, we get $(e - f) \\cdot e = 1 - e \\cdot f$ and $(e - f) \\cdot f = e \\cdot f - 1$.\n9. Hence $(e - f) \\cdot w = (1 - t) c \\, (1 - e \\cdot f) - t b \\, (1 - e \\cdot f) = \\left( (1 - t) c - t b \\right) (1 - e \\cdot f)$.\n10. The vectors $u$ and $v$ are [[linearly independent|0CRMGPAH83]], because the vertices are not collinear. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]] and its case of equality, $u \\cdot v \u003c c b$. So $e \\cdot f \u003c 1$.\n11. By steps 7, 9 and 10, $\\angle BAD = \\angle DAC$ if and only if $(1 - t) c = t b$.\n12. Assume that $\\angle BAD = \\angle DAC$. Then $(1 - t) c = t b$. If $t = 1$, this says $0 = b$, which is false. So $t \u003c 1$, and $d(D, C) = (1 - t) a \u003e 0$.\n13. By steps 3 and 12, $\\dfrac{d(B, D)}{d(D, C)} = \\dfrac{t}{1 - t} = \\dfrac{c}{b} = \\dfrac{d(A, B)}{d(A, C)}$. This proves the first claim.\n14. Conversely, assume that $\\dfrac{d(B, D)}{d(D, C)} = \\dfrac{d(A, B)}{d(A, C)}$. The quotient on the left is defined, so $d(D, C) \u003e 0$ and $t \u003c 1$.\n15. By step 3, the assumption says $\\dfrac{t}{1 - t} = \\dfrac{c}{b}$, that is $t b = (1 - t) c$. By step 11, $\\angle BAD = \\angle DAC$.\n\n## Idea\n\nThe ray $AD$ makes equal angles with $AB$ and $AC$ exactly when it has equal dot products with the two unit vectors along them. Writing $D$ on the segment $BC$ turns this into one linear equation in the position of $D$. Its solution divides $BC$ in the ratio $d(A, B) : d(A, C)$.\n",
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        "url": "https://raw.org/book/linear-algebra/bisector-of-two-vectors/"
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        "url": "https://en.wikipedia.org/wiki/Angle_bisector_theorem"
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