Angle Bisector Theorem is the habit of using that a bisector divides the opposite side in the ratio of the adjacent sides. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
The internal bisector of an angle of a triangle divides the opposite side into segments proportional to the two sides forming that angle.
Why It Works
1
Let the bisector of $\angle A$ in $\triangle ABC$ meet $BC$ at $D$, so $\angle BAD = \angle DAC$.
2
Triangles $\triangle ABD$ and $\triangle ACD$ share the same height from $A$ to line $BC$, so their areas are in the ratio of their bases: $\dfrac{[\triangle ABD]}{[\triangle ACD]} = \dfrac{BD}{DC}$.
3
Using the other base, $\dfrac{[\triangle ABD]}{[\triangle ACD]} = \dfrac{\tfrac12\,AB\cdot AD\sin\angle BAD}{\tfrac12\,AC\cdot AD\sin\angle DAC} = \dfrac{AB}{AC}$, since the bisected angles are equal so the sines cancel.
4
Equating the two expressions for the area ratio gives $\dfrac{BD}{DC} = \dfrac{AB}{AC}$.
Worked Examples
Example 1
In $\triangle ABC$, $AB = 6$ and $AC = 9$. The internal bisector of $\angle A$ meets $BC$ at $D$. If $BC = 10$, find $BD$.
1
By the angle bisector theorem, $\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{6}{9} = \dfrac{2}{3}$.
2
Write $BD = 2k$ and $DC = 3k$; then $BD + DC = 5k = BC = 10$, so $k = 2$.
3
Therefore $BD = 2k = 4$.
Answer:
$BD = 4$
Warm-up
In $\triangle ABC$, $AB = 10$ and $AC = 14$. The internal bisector of $\angle A$ meets $BC$ at $D$, and $BC = 12$. Find $BD$ and $DC$.
Write $BD = 5k$, $DC = 7k$; then $5k + 7k = 12k = BC = 12$, so $k = 1$.
3
Therefore $BD = 5$ and $DC = 7$.
Answer:
$BD = 5$, $DC = 7$.
Contest level
In $\triangle ABC$ the internal bisector of $\angle A$ meets $BC$ at $D$ with $BD = 4$ and $DC = 6$. If $AB = 6$, find $AC$.
1
Angle-bisector theorem: $\dfrac{BD}{DC} = \dfrac{AB}{AC}$, so $\dfrac{4}{6} = \dfrac{6}{AC}$.
2
Cross-multiply: $4\cdot AC = 6\cdot 6 = 36$.
3
Therefore $AC = 9$.
Answer:
$AC = 9$
Olympiad / Challenge
In $\triangle ABC$ with $BC = a$, $CA = b$, $AB = c$, the internal bisector from $A$ meets $BC$ at $D$, and the incenter $I$ lies on $AD$. Prove $\dfrac{AI}{ID} = \dfrac{b + c}{a}$.
1
By the angle-bisector theorem at $A$, $\dfrac{BD}{DC} = \dfrac{c}{b}$, so $BD = \dfrac{c}{b + c}\cdot a = \dfrac{ac}{b + c}$.
2
The incenter $I$ is also on the bisector from $B$; in $\triangle ABD$ the segment $BI$ bisects $\angle ABD$, so by the angle-bisector theorem in that triangle $\dfrac{AI}{ID} = \dfrac{AB}{BD} = \dfrac{c}{BD}$.
(For instance, with $a = 6$, $b = 5$, $c = 4$ this gives $AI:ID = 9:6 = 3:2$.)
Answer:
$\dfrac{AI}{ID} = \dfrac{b + c}{a}$.
Going Deeper
Generalization: there is a twin for the EXTERNAL bisector — it meets line $BC$ at the point $D'$ dividing $BC$ EXTERNALLY in the same ratio $\dfrac{BD'}{D'C} = \dfrac{AB}{AC}$. The internal foot $D$ and external foot $D'$ together with $B, C$ form a harmonic range, the bridge to the Apollonius circle (locus of points with $\dfrac{PB}{PC} = \dfrac{c}{b}$).
Where it appears: this is the first move on essentially every problem involving an angle bisector or an incenter — it converts the bisector into a side ratio, after which mass points or Stewart's theorem finish the job (the incenter ratio $AI:ID = (b+c):a$ in the Olympiad example is a direct consequence).
Pitfall: the ratio is $\dfrac{BD}{DC} = \dfrac{AB}{AC}$ — the segment is proportional to the ADJACENT side ($BD$ to $AB$, since $B$ is shared). Flipping it to $\dfrac{BD}{DC} = \dfrac{AC}{AB}$ is the standard error. Also remember this is the INTERNAL bisector; the external bisector gives an external division, not the same point.
Spot the Signal
Use it when an angle bisector meets the opposite side and ratios or lengths are wanted.
You can describe the hard part as using that a bisector divides the opposite side in the ratio of the adjacent sides, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by writing $\frac{BD}{DC} = \frac{AB}{AC}$ and solving for the unknown segment.
Name the relation that makes Angle Bisector Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Applying it to a median or an altitude instead of a genuine angle bisector.
Applying Angle Bisector Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
The bisector shares the opposite side in the ratio of the sides that form the angle.
Try it on:
Find a segment ratio or length created by an angle bisector.