Finding an angle between side and a segment from specified point inside an equilateral triangle Here is the question:



*

*$\overset{\Delta}{ABC}$ is an equilateral triangle.

*D is a point inside triangle.

*$m(\widehat{BAD})=12^\circ$

*$m(\widehat{DBA})=6^\circ$

*$m(\widehat{ACD})=x=?$


I managed to reduce answer to $\cot x=\csc 48^\circ+\sqrt 3$ with the help of Trig-Ceva Theorem, but i can't find x without a calculator. I suspect that there is an elementary (but complex) method to solve this but i'm not sure.
[The answer i'm trying to reach is $18^\circ$]
 A: This problem is actually part of a regular $30$-gon, as shown in the following figure. For a proof that three diagonal lines meet at a single point, refer to Section $3$ and Table $4$ (page $13$) of http://www-math.mit.edu/%7Epoonen/papers/ngon.pdf. (Unfortunately, this is not so easy. The following lemma from Section $2$ (pages $4-5$) helps.)
Lemma. Let $A$, $B$, $C$, $D$, $E$, $F$ be six distinct points in order on a circle. The three chords $AD$, $BE$, $CF$ meet at a single point if and only if $AF\cdot BC\cdot DE=CD\cdot EF\cdot AB$.
A: I document two answers to this question; one is trigonometric answer i came up with, other is geometric answer i found on the net. Geometric answer is in another language, so i translated it to English instead of linking it.
First:

For easier reference, some known formulas:
$$2\sin x\sin y=\cos(x-y)-\cos(x+y)$$
$$2\sin x\cos y=\sin(x+y)+\sin(x-y)$$
$$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$

Second:

Proof that $\sin54^\circ\sin18^\circ=1/4$:
Insert $\sin72^\circ=2\sin36^\circ\sin54^\circ$ into $\sin36^\circ=2\sin18^\circ\sin72^\circ$ to get $\sin54^\circ\sin18^\circ=1/4$.

I solved this question with the Trig-Ceva theorem. From the theorem:
$$
\frac{\sin12^\circ}{\sin48^\circ}\frac{\sin54^\circ}{\sin6^\circ}\frac{\sin\alpha}{\sin(60^\circ-\alpha)}=1
$$
And we reach:
$$
2\cos6^\circ\sin54^\circ\sin\alpha=\sin48^\circ\sin(60^\circ-\alpha)\tag{1}
$$
Since $2\sin48^\circ\sin42^\circ=\cos6^\circ$ and $\sin54^\circ\sin18^\circ=1/4$, we can make advance guessing and find $\alpha=18^\circ$ from $(1)$. Else, we need to tamper the equation $(1)$ more:
$$
\frac{\sin(60^\circ-\alpha)}{\sin\alpha}=\frac{\sqrt3/2+\sin48^\circ}{\sin48^\circ}\\
\frac{\sqrt3/2\cos\alpha-1/2\sin\alpha}{\sin\alpha}=\sqrt3/2\csc48^\circ+1
$$
And we get:
$$
\cot\alpha=\csc48^\circ+\sqrt3\tag{2}
$$
Now we can prove that $\cot18^\circ=\csc48^\circ+\sqrt3$ by tracing the steps back to equation $(1)$ from equation $(2)$. After that; because we know, since $\cot\alpha$ is one-to-one for $0<\alpha<180^\circ$, there is only one $\alpha$ value to maintain equation $(2)$, we find that $\alpha=18^\circ$. $\blacksquare$

Here is the second proof:

Draw angle bisector of $\widehat{BAD}$, intersecting $[BD]$ in $E$. Draw $[CE]$. Since $\overset{\Delta}{ABE}$ is isosceles, $[CE]$ is the axis of symmetry of $\overset{\Delta}{ABC}$, making $m(\widehat{ECB})=30^\circ$. Draw $[BF]$ such that $m(\widehat{FBD})=6^\circ$ and $|BF|=|AB|$. Draw $[AF]$ and $[CF]$. Since $\overset{\Delta}{BCF}$ is isosceles and $m(\widehat{CBF})=48^\circ$, then $m(\widehat{FCB})=66^\circ$, therefore $m(\widehat{FCA})=6^\circ$. Draw $[DF]$. Since $\overset{\Delta}{ABF}$ is isosceles, $[BD]$ is the axis of symmetry of $\overset{\Delta}{ABF}$, making $m(\widehat{DFB})=12^\circ$. Extend $[AD]$; since $[CE]$ is the axis of symmetry of $\overset{\Delta}{ABC}$ and $m(\widehat{FBA})=m(\widehat{BAD})=12^\circ$, extension of $[AD]$ meet at the intersection of two segments $G$. Since $\widehat{FGA}$ is an exterior angle of $\overset{\Delta}{ABG}$, $m(\widehat{FGA})=24^\circ$. Since $\widehat{FDA}$ is an exterior angle of $\overset{\Delta}{DFG}$, $m(\widehat{FDA})=36^\circ$. Since $m(\widehat{FCG})=36^\circ$ opposite angles of $\overset{\square}{CFDG}$ are supplementary, making this a cyclic quadrilateral, meaning its four points are on a circle. Since $\widehat{FGD}$ and $\widehat{FCD}$ both looking at the same arc in this circle, $m(\widehat{FCD})=24^\circ$. And then, finally, we reach $\alpha=18^\circ$. $\blacksquare$
A: The following proof (for the general problem - $2\widehat{ACD}=3\widehat{BAD}=6\widehat{ABD}=6\alpha$) I saw on a facebook Peruvian group:
Take $E$ on $AB|DE=BE=AD$, draw the perpendicular bisector of $BC$, take $F$ on it so that $FD=AB$; clearly $EDF$ is an equilateral triangle, since $D$ is the circumcenter of $\triangle AFE$, and $\widehat{AFD}=\widehat{FAD}=30^\circ-2\alpha$ ( 1 ).
$\widehat{AFC}=\widehat{AFB}=120^\circ-\alpha$ ( 2 ), since $\widehat{ABF}=30^\circ+\alpha, \widehat{FAB}=30^\circ$, but also $\widehat{ADC}=120^\circ-\alpha$, since $\widehat{ACD}+\widehat{CAD}=60^\circ+\alpha$, consequently $DACF$ is cyclic and $\widehat{ACD}=\widehat{AFD}$, i.e. $3\alpha=30^\circ-2\alpha$, implying $5\alpha=30^\circ, \alpha=6^\circ$.
For the given problem the proof is slightly different:$\widehat{ACF}=\widehat{ABF}=36^\circ$ ( from symmetry ), while $\widehat{AFB}=\widehat{FAB}=18^\circ$, so $\widehat{ADF}=144^\circ$, again $FDAC$ is cyclic, hence $\widehat{ACD}=\widehat{AFD}=18^\circ$.

