Angle Bisector Theorem/Exterior Angle

Theorem

Let $\triangle ABC$ be a triangle.

Let $AB$ be produced past $A$ to $D$.

Let the external angle $CAD$ be bisected by $AE$ where $BE$ is $BC$ produced.

Then:

$BE : EC = AB : AC$

Proof

Construct $CF$ parallel to the angle bisector $AE$.

By Parallelism implies Equal Corresponding Angles:

$\angle AFC = \angle DAE$

By Parallelism implies Equal Alternate Angles:

$\angle ACF = \angle CAE$

Given that $\angle DAC$ is bisected:

$\angle DAE = \angle CAE$

By Common Notion $1$:

$\angle AFC = \angle ACF$

By Triangle with Two Equal Angles is Isosceles:

$\triangle AFC$ is isosceles

By the definition of isosceles triangle:

$AF = AC$

By Parallelism implies Equal Corresponding Angles:

$\angle BAE = \angle BFC$

$\angle ABC$ is shared.

By Triangles with Two Equal Angles are Similar:

$\triangle BFC \sim \triangle BAE$

By the definition of similar triangles:

$AB : BE = AF : CE$

Substituting for $AF$:

$AB : BE = AC : CE$

Rearrange to give:

$AB : AC = BE : EC$

$\blacksquare$

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): angle bisector theorem