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.