Angle Bisector Proportionality Theorem

A proof that the bisector of a triangle's angle divides the opposite side in the ratio of the two adjacent sides, BD:DC = AB:AC, via a line drawn through the far vertex parallel to the bisector.

Keywords: angle bisector theorem, proportionality, parallel line, similar triangles, isosceles triangle

Prerequisites: parallel-lines-intro, isosceles-triangle-proof, triangle-similarity · Difficulty: intermediate

The bisector of an angle of a triangle divides the opposite side into segments proportional to the two adjacent sides. With the bisector from $A$ meeting $BC$ at $D$,

Strategy: through $C$, draw a line parallel to the bisector $AD$ and extend side $BA$ to meet it at $E$. The parallel makes triangle $ACE$ isosceles ($AE=AC$), and in triangle $BCE$ the parallel cuts the sides in the ratio $BD:DC=BA:AE=c:b$.

Step 1: The Bisector Meets $BC$ at $D$

In triangle $ABC$, the bisector of angle $A$ meets the opposite side $BC$ at $D$, splitting $\angle A$ into two equal halves.

Step 2: $BD:DC=AB:AC=c:b$

Claim. The bisector from $A$ divides $BC$ in the ratio of the two adjacent sides,

We prove it by drawing, through $C$, a line parallel to the bisector.

Step 3: Draw a Parallel to the Bisector

Through $C$, draw the line parallel to the bisector $AD$. Extending side $BA$, the two meet at a point $E$ beyond $A$, so $CE\parallel AD$.

Step 4: $AE=AC$

Since $CE\parallel AD$, corresponding angles give $\angle AEC=\angle BAD$ and alternate angles give $\angle ACE=\angle DAC$. The bisector makes $\angle BAD=\angle DAC$, so $\angle AEC=\angle ACE$: triangle $ACE$ is isosceles and $AE=AC$.

Step 5: $BD:DC=BA:AE$

In triangle $BCE$, the bisector $AD$ is parallel to side $CE$, so it cuts the other two sides proportionally — equivalently $△BAD\sim △BEC$:

Since $AE=AC$,

Step 6: The Proportionality Theorem

The bisector from $A$ divides $BC$ in the ratio of the adjacent sides,

The same ratio also follows from comparing the areas of triangles $ABD$ and $ACD$, which share the apex $A$ — an idea that returns when proving Ceva's theorem in a later chapter.

The bisector from $A$ divides the opposite side in the ratio of the adjacent sides,

This is the form of the Angle Bisector Theorem the incenter coordinate formula relies on. The same ratio also follows from an area argument — comparing triangles $ABD$ and $ACD$, which share the apex $A$ — an idea that returns when proving Ceva's theorem in a later chapter. ∎