---
tag: G26V4F9BY5
kind: proof
name: Proof of the angle bisector theorem and its converse with unit vectors
msc: [51M04]
proves: 0859R2V6ES
uses: [0CTNFGYMM2, 0GJAJC6TNZ, 0NPPGHQNXQ, 0AFNNTRTJE, VQ4Q2V45RP, 0XKBKGTCPC, 0S52TNNTH9, 0CRMGPAH83, 0RFWW56Y22, 0CWDVN132S]
sources:
  - url: https://raw.org/book/linear-algebra/bisector-of-two-vectors/
    locator: Algebraic Proof
  - url: https://en.wikipedia.org/wiki/Angle_bisector_theorem
    locator: Theorem
---

## Proof

We reduce both directions to one identity for dot products with unit vectors.

1. 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 $a, b, c > 0$.
2. Since $D$ is a point of the [[segment|VQ4Q2V45RP]] $BC$, there is $t \in [0, 1]$ with $D = (1 - t) B + t C$.
3. Then $D - B = t (C - B)$ and $D - C = (1 - t)(B - C)$. So $d(B, D) = t a$ and $d(D, C) = (1 - t) a$.
4. 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.
5. 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]] $1$.
6. 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 $\cos \angle DAC = \dfrac{w \cdot v}{\norm{w} \, b} = \dfrac{f \cdot w}{\norm{w}}$.
7. 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 $(e - f) \cdot w = 0$.
8. 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 $(e - f) \cdot f = e \cdot f - 1$.
9. Hence $(e - f) \cdot w = (1 - t) c \, (1 - e \cdot f) - t b \, (1 - e \cdot f) = \left( (1 - t) c - t b \right) (1 - e \cdot f)$.
10. 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, $u \cdot v < c b$. So $e \cdot f < 1$.
11. By steps 7, 9 and 10, $\angle BAD = \angle DAC$ if and only if $(1 - t) c = t b$.
12. Assume that $\angle BAD = \angle DAC$. Then $(1 - t) c = t b$. If $t = 1$, this says $0 = b$, which is false. So $t < 1$, and $d(D, C) = (1 - t) a > 0$.
13. 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.
14. 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 $d(D, C) > 0$ and $t < 1$.
15. By step 3, the assumption says $\dfrac{t}{1 - t} = \dfrac{c}{b}$, that is $t b = (1 - t) c$. By step 11, $\angle BAD = \angle DAC$.

## Idea

The 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 $d(A, B) : d(A, C)$.