$\triangle ABC$ with a point $D$ inside has $\angle BAD=114^\circ$, $\angle DAC=6^\circ$, $\angle ACD=12^\circ$, and $\angle DCB=18^\circ$. 
Let $ABC$ be a triangle with a point $D$ inside.  Suppose that $\angle BAD=114^\circ$, $\angle  DAC=6^\circ$, $\angle ACD=12^\circ$ and $\angle DCB=18^\circ$.  Show that $$\frac{BD}{AB}=\sqrt2.$$


I am requesting a geometric proof (with as little trigonometry as possible). A completely geometric proof would be most appreciated.  I have a trigonometric proof below.

Trigonometric Proof
Wlog, let $AB=1$.  Note that $\angle ABC=\angle ACB=30^\circ$, so $AC=1$.  Then by law of sines on $\triangle ACD$, $$AD=\frac{\sin 12^\circ}{\sin 18^\circ}.$$  By law of cosines on $\triangle ABD$,
$$BD^2=1^2+\frac{\sin^212^\circ}{\sin^2{18^\circ}}-2\frac{\sin 12^\circ}{\sin 18^\circ}\cos 114^\circ.$$
As $\cos 114^\circ=-\sin24^\circ$, we get
$$BD^2=2+\frac{-\sin^218^\circ+\sin^212^\circ+2\sin12^\circ\sin18^\circ\sin 24^\circ}{\sin^218^\circ}.$$
Then from the identities $\sin^2\alpha-\sin^2\beta=\sin(\alpha-\beta)\sin(\alpha+\beta)$ and $\sin(2\alpha)=2\sin\alpha\cos\alpha$, we have
$$BD^2=2+\frac{-\sin 6^\circ\sin 30^\circ+4\sin 6^\circ\cos 6^\circ \sin 18^\circ\sin24^\circ}{\sin^218^\circ}.$$
Because $\sin 30^\circ=\frac12$, we conclude that $BD=\sqrt{2}$ if we can prove
$$8\cos 6^\circ \sin 18^\circ \sin 24^\circ=1.$$
This is true because by the identity $2\sin\alpha\cos\beta=\sin({\alpha+\beta})+\sin(\alpha-\beta)$, we have
$$2\sin 24^\circ \cos 6^\circ =\sin 30^\circ+\sin 18^\circ.$$
Since $\sin 30^\circ=\frac12$, we obtain
$$8\cos 6^\circ \sin 18^\circ \sin 24^\circ =2\sin 18^\circ +4\sin^218^\circ=1,$$
noting that $\sin 18^\circ=\frac{\sqrt5-1}{4}$.

Attempt at Geometric Proof
I discovered something that might be useful.  Construct  the points $E$ and $G$ outside $\triangle ABC$ so that $\triangle EBA$ and $\triangle GAC$ are similar to $\triangle ABC$ (see the figure below).  Clearly, $EAG$ is a straight line parallel to $BC$.  Let $F$ and $H$ be the points corresponding to $D$ in $\triangle EBA$ and $\triangle GAC$, respectively (that is, $\angle FAB=\angle DCB=\angle HCA$ and $\angle FAE=\angle DCA=\angle HCG$). Then $\triangle FBD$  and $\triangle HDC$ are isosceles triangles similar to $\triangle ABC$, and $\square AFDH$ is a parallelogram.  I haven't been able to do anything further than this without trigonometry.

Here is a bit more attempt.  If $M$ is the reflection of $A$ wrt $BC$, then through the use of trigonometric version of Ceva's thm, I can prove that $\angle AMD=42^\circ$ and $\angle CMD=18^\circ$.  Not sure how to prove this with just geometry.  But this result may be useful.  (Although we can use law of sines on $\triangle MCD$ to get $MD$ and then use law of cosines on $\triangle BMD$ to get $BD$ in terms of $AB$ too.  But this is still a heavily trigonometric solution, even if the algebra is less complicated than the one I wrote above.)

I have a few more observations.  They may be useless.  Let $D'$ be the point obtained by reflecting $D$ across the perpendicular bisector of $BC$.  Draw a regular pentagon $ADKK'D'$.  Geogebra tells me that $\angle ABK=54^\circ$ and $\angle AKB=48^\circ$.  This can be proven using trigonometry, although a geometric proof should exist.  But it is easy to show that $KD\perp CD$ and $K'D'\perp BD'$.

In all of my attempts, I always ended up with one of the following two trigonometric identities:
$$\cos 6^\circ \sin 18^\circ \sin 24^\circ=1/8,$$
$$\cos 36^\circ-\sin18^\circ =1/2.$$
(Of course these identities are equivalent.)  I think a geometric proof will need an appearance of a regular pentagon and probably an equilateral triangle, and maybe a square.

This is an in-complete proof because I am stuck from step 8 onward.


*

*Draw Z on BC such that $\angle BAZ = 90^0$.


*Let CZ = 1. Then AZ = 1 because $\triangle ZAC$ is isosceles.


*Because $\angle ABC = 30^0$, $AC = AB = \sqrt 3$.


*Construct the blue circle (centered at B, radius $= BA = \sqrt 3$.


*From D draw the tangent to circle (B) touching it at X. Then $\angle BXD = 90^0$,


*Let Y be the midpoint of BC.


*Draw CV // YX. By intercept theorem, BX = XV. Together with the finding in (5), we can say that DX is the perpendicular bisector of BV.


*Draw the circle passing through B, D, V. It will cut the red circle (A) at some point U.
[Another way to let X be the center of the dotted circle and prove that D is a con-cyclic point of that circle.]
If we can show that X is the center of the dotted circle, then $\triangle DBV$ is 45-45-90. In turn, $\triangle XBD$ is also 45-45-90. Consequently, required result follows from the fact that $BD = \sqrt 2 \times \sqrt 3$.
