Angle bisector theorem

In the above diagram, use the law of sines on triangles ABD and ACD:

${\displaystyle {\frac {|AB|}{|BD|}}={\frac {\sin \angle ADB}{\sin \angle DAB}}}$

(1)

${\displaystyle {\frac {|AC|}{|CD|}}={\frac {\sin \angle ADC}{\sin \angle DAC}}}$

(2)

Angles ∠ ADB and ∠ ADC form a linear pair, that is, they are adjacent supplementary angles. Since supplementary angles have equal sines,

${\displaystyle {\sin \angle ADB}={\sin \angle ADC}.}$

Angles ∠ DAB and ∠ DAC are equal. Therefore, the right hand sides of equations (1) and (2) are equal, so their left hand sides must also be equal.

${\displaystyle {\frac {|BD|}{|CD|}}={\frac {|AB|}{|AC|}},}$

which is the angle bisector theorem.

If angles ∠ DAB and ∠ DAC are unequal, equations (1) and (2) can be re-written as:

${\displaystyle {{\frac {|AB|}{|BD|}}\sin \angle \ DAB=\sin \angle ADB},}$

${\displaystyle {{\frac {|AC|}{|CD|}}\sin \angle \ DAC=\sin \angle ADC}.}$

Angles ∠ ADB and ∠ ADC are still supplementary, so the right hand sides of these equations are still equal, so we obtain:

${\displaystyle {{\frac {|AB|}{|BD|}}\sin \angle \ DAB={\frac {|AC|}{|CD|}}\sin \angle \ DAC},}$

which rearranges to the "generalized" version of the theorem.

Let D be a point on the line BC, not equal to B or C and such that AD is not an altitude of triangle ABC.

Let B₁ be the base (foot) of the altitude in the triangle ABD through B and let C₁ be the base of the altitude in the triangle ACD through C. Then, if D is strictly between B and C, one and only one of B₁ or C₁ lies inside triangle ABC and it can be assumed without loss of generality that B₁ does. This case is depicted in the adjacent diagram. If D lies outside of segment BC, then neither B₁ nor C₁ lies inside the triangle.

∠ DB₁B and ∠ DC₁C are right angles, while the angles ∠ B₁DB and ∠ C₁DC are congruent if D lies on the segment BC (that is, between B and C) and they are identical in the other cases being considered, so the triangles DB₁B and DC₁C are similar (AAA), which implies that

${\displaystyle {\frac {|BD|}{|CD|}}={\frac {|BB_{1}|}{|CC_{1}|}}={\frac {|AB|\sin \angle BAD}{|AC|\sin \angle CAD}}.}$

If D is the foot of an altitude, then,

${\displaystyle {\frac {|BD|}{|AB|}}=\sin \angle \ BAD{\text{ and }}{\frac {|CD|}{|AC|}}=\sin \angle \ DAC,}$

and the generalized form follows.

${\displaystyle \alpha ={\tfrac {\angle BAC}{2}}=\angle BAD=\angle CAD}$

A quick proof can be obtained by looking at the ratio of the areas of the two triangles ${\displaystyle \triangle BAD}$ and ${\displaystyle \triangle CAD}$, which are created by the angle bisector in ${\displaystyle A}$. Computing those areas twice using different formulas, that is ${\displaystyle {\tfrac {1}{2}}gh}$ with base ${\displaystyle g}$ and altitude ${\displaystyle h}$ and ${\displaystyle {\tfrac {1}{2}}ab\sin(\gamma )}$ with sides ${\displaystyle a}$, ${\displaystyle b}$ and their enclosed angle ${\displaystyle \gamma }$, will yield the desired result.

Let ${\displaystyle h}$ denote the height of the triangles on base ${\displaystyle BC}$ and ${\displaystyle \alpha }$ be half of the angle in ${\displaystyle A}$. Then

${\displaystyle {\frac {|\triangle ABD|}{|\triangle ACD|}}={\frac {{\frac {1}{2}}|BD|h}{{\frac {1}{2}}|CD|h}}={\frac {|BD|}{|CD|}}}$

and

${\displaystyle {\frac {|\triangle ABD|}{|\triangle ACD|}}={\frac {{\frac {1}{2}}|AB||AD|\sin(\alpha )}{{\frac {1}{2}}|AC||AD|\sin(\alpha )}}={\frac {|AB|}{|AC|}}}$

yields

${\displaystyle {\frac {|BD|}{|CD|}}={\frac {|AB|}{|AC|}}.}$