In $\triangle ABC$ with an internal point $D$, $\angle BCA=2\angle BAC$ and $AB=12$, find $CD$ As the title says, the question is to solve for the length $CD$ in the figure below. I found problem online in a Japanese forum, however it had no replies/solutions, so I've decided to share it here to see what kind of approaches are possible for this problem.
Please do not mind the Japanese text in the image, all the relevant information can already be seen, here, $AB=12$, $\angle BCA=2\angle BAC$ and $CD$ is the unknown. I'll share my approach as an answer below, please share your answers too!

 A: In this case, trigonometric solution is easier than the geometric method.
From sin rule,
$$\frac{BC}{\sin\alpha}=\frac{12}{\sin2\alpha}=\frac{12}{2\sin\alpha\cos\alpha}\implies BC\cos\alpha=6.$$
And $CD=BC\cos\alpha=6$.
A: This is my approach:

1.) Extend $BD$ such that it meets $E$ on segment $AC$ such that $\triangle BEC$ is an isosceles triangle. Therefore $BC=CE$ and $BD=DE$. Draw a line from $D$ onto point $F$ on $AB$ such that
$DF \parallel EA$.
2.) Via the AAA property, we can see that $\triangle BFD$ is similar to $\triangle BAE$. We can also see, since $DF \parallel EA$ and $\angle FAC=\angle DCA=\alpha$, that $DC=AF$. Now, via the basic proportionality theorem, we can say that:
$$\frac{BF}{FA}=\frac{BD}{DE}=1$$
$$BF=FA$$
Since $AB=12$, we can infer that $BF=FA=DC=6$
A: 
Let O be the circumcenter of the $\triangle BAC$ if we take the angles marked with '.' equal to $\alpha$. Then by simple theorems, you can find that $\angle BOC = 2\alpha$ and $\angle OBC=90-\alpha$. Because $\angle DBC = 90-\alpha$ (already found) D lies on the radius $OB$
Now finally draw a perpendicular from O intersecting at X with AB
$BX = 6$, Because a perp. segment from the center of a circle divides a chord in half
Now consider $\triangle BOX$ and $\triangle ODC$
By S.A.A, $\triangle BOX \cong \triangle ODC$
Because the respective sides of congruent triangles are equal,
$DC=BX=6$
