Angle bisector theorem

Builds on Triangle Angle at a point Euclidean distance

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.

Reading path

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Tag	`0859R2V6ES`
URI	`m2://0859R2V6ES`
Kind	statement, theorem
Writing	W1 stated
MSC	51M04
Status	This copy of the site has no build data. Run `m2 status 0859R2V6ES`.
Uses	Triangle, Angle at a point, Euclidean distance
Same as	Wikidata Q925854 Wikipedia Angle bisector theorem

Angle bisector theorem

Statement

Remarks

History

Reading path

Further reading

What this page needs for W2

What links here