Angle bisector theorem

Statement

Let $ABC$ be a triangle in the Euclidean space $\mathbb{R}^n$, and let $d$ be the Euclidean distance. Let $D$ be a point of the segment $BC$, and let $\angle$ denote the angle at a point. If $\angle BAD = \angle DAC$, then $$\frac{d(B, D)}{d(D, C)} = \frac{d(A, B)}{d(A, C)}.$$ Conversely, if $D$ is a point of the segment $BC$ with this equality, then $\angle BAD = \angle DAC$.

Remarks

In words, the bisector of the angle at $A$ cuts the opposite side in the ratio of the two adjacent sides. In an isosceles triangle with $d(A, B) = d(A, C)$, the bisector from $A$ meets $BC$ at its midpoint. A similar statement holds for the bisector of the exterior angle at $A$, which meets the line $BC$ outside the segment.

History

The theorem is Proposition 3 of Book VI of Euclid's Elements.

Explanations

L1 human

Take the triangle with $d(A, B) = 6$ and $d(A, C) = 3$. The bisector of the angle at $A$ cuts $BC$ into two parts. The part next to $B$ is twice as long as the part next to $C$.