Proof of Angle Bisector Theorem

Angle Bisector Theorem

Angle Bisector Theorem uses angle bisectors of triangles and states a relation between lengths of segments along the triangle.

Formal Definition

Given any triangle $ABC$ with point $D$ on $AC$ such that $\angle BAD= \angle CAD$

Angle Bisector Theorem States that $\frac { BD}{ DC }=\frac { AB }{ AC}$

Proof

Apply law of sines on $\triangle ABD$ and $\triangle ACD$ to get $\frac {AB} {BD} = \frac {\sin \angle BDA} {\sin \angle BAD}$ and $\frac {AC} {DC} = \frac {\sin \angle ADC} {\sin \angle DAC}$

Since $\angle BDA$ is supplementary to $\angle ADC$,

$\sin{ (ADC )}= \sin{ (BDA)}$

Also, $\angle BAD$ and $\angle DAC$ are equal, thus $\frac{AB}{BD}=\frac{AC}{DC}$ Re-arrange to get $\frac { BD}{ DC }=\frac { AB }{ AC}$

** Generalized Angle Bisector Theorem**

Given any triangle $ABC$ with point $D$ on $AC$. Then Generalized Angle Bisector Theorem states that $\frac {BD}{DC}=\frac {AB \sin \angle DAB}{AC \sin \angle DAC}$

The internal angle bisector theorem says that given any triangle ABC with the foot of the angle bisector of A at D, $\frac{AC}{CD}=\frac{AB}{BD}$

To prove this, we can construct a line from point $B$ parallel to $AC$. Let's call the point at which it intersects $AD$ point $E$.

Since $BE \parallel AC$, $\angle CAE = \angle BEA$ by alternate interior angles. But $\angle CAE$ is also equal to $\angle BAE$ by construction. This means that $\Delta ABE$ is isosceles, and $AB = BE$.

Now look at triangles $ADC$ and $EDB$. We already know that $\angle CAE = \angle BEA$, and we can also see that $\angle ADC = \angle EDB$ by vertical angles. Thus, $\Delta ADC \sim \Delta EDB$. From this similarity we can write down that $\frac{AC}{BE}=\frac{CD}{BD}$ Substituting in $AB = BE$ gives us $\frac{AC}{AB} = \frac{CD}{BD}$. Rearranging this, we get $\frac{AC}{CD}=\frac{AB}{BD}$ Which is the angle bisector theorem.

Alternatively, we can also prove this using Law of Sines and noting that $\sin ADB=\sin\left(180-ADC\right)=\sin\left(ADC\right)$ Note that the converse of the angle bisector theorem is also true, so if $\frac{AC}{CD}=\frac{AB}{BD}$, $AD$ bisects $\angle BAC$. This is a very useful way of proving that a line is an angle bisector, or converting angle equality into side ratios without similar triangles