In triangle $ABC$, $AB = 6$, $AC = 3$, and $\angle BAC = 120^{\circ}$. $D$ is the intersection of the angle bisector of $\angle BAC$ and line $BC$. $E$ is the intersection of the external angle bisector of $\angle BAC$ and line $BC$. We need to find $BC$, the ratio $BD:CD$, the ratio $BE:CE$, and $DE$.

1. Problem Description

In triangle $ABC$, $AB = 6$, $AC = 3$, and $\angle BAC = 120^{\circ}$. $D$ is the intersection of the angle bisector of $\angle BAC$ and line $BC$. $E$ is the intersection of the external angle bisector of $\angle BAC$ and line $BC$. We need to find $BC$, the ratio $BD:CD$, the ratio $BE:CE$, and $DE$.

2. Solution Steps

First, we use the Law of Cosines to find $BC$:

$BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos{120^{\circ}}$

$BC^2 = 6^2 + 3^2 - 2(6)(3)(-\frac{1}{2})$

$BC^2 = 36 + 9 + 18$

$BC^2 = 63$

$BC = \sqrt{63} = 3\sqrt{7}$

By the Angle Bisector Theorem, $\frac{BD}{CD} = \frac{AB}{AC} = \frac{6}{3} = 2$.

So, $BD:CD = 2:1$.

By the External Angle Bisector Theorem, $\frac{BE}{CE} = \frac{AB}{AC} = \frac{6}{3} = 2$.

So, $BE:CE = 2:1$.

Since $BD:CD = 2:1$, we have $BD = 2CD$.

Also, $BD + CD = BC = 3\sqrt{7}$.

Substituting $BD = 2CD$, we get $2CD + CD = 3\sqrt{7}$, so $3CD = 3\sqrt{7}$, which gives $CD = \sqrt{7}$.

Then $BD = 2CD = 2\sqrt{7}$.

Since $BE:CE = 2:1$, we have $BE = 2CE$.

Also, $BE - CE = BC = 3\sqrt{7}$.

Substituting $BE = 2CE$, we get $2CE - CE = 3\sqrt{7}$, so $CE = 3\sqrt{7}$.

Then $BE = 2CE = 6\sqrt{7}$.

Now, we want to find $DE = DC + CE = \sqrt{7} + 3\sqrt{7} = 4\sqrt{7}$.

3. Final Answer

$BC = 3\sqrt{7}$

$BD:CD = 2:1$

$BE:CE = 2:1$

$DE = 4\sqrt{7}$