Find $\angle{CAD}$ in equilateral triangle $\triangle{ABC}$ for $D$ inside with $\angle{ABD}=18^{\circ}, \angle{BCD}=12^{\circ}$ 
This problem is feasible to solve using trigonometric functions. I am looking for pure geometric solution as usual. Thanks...


Let $E$ be the circumcenter of $\triangle{BCD}$, so
$BE=CE=DE, \angle{BED}=2\angle{BCD}=2*12^{\circ}=24^{\circ} \\
\implies \angle{EBD}=\angle{BDE}=\dfrac{180^{\circ}-24^{\circ}}{2}=78^{\circ} \\
\implies \angle{EBC}=\angle{ECB}=78^{\circ}-42^{\circ}=36^{\circ} \\
\implies \angle{BEC}=108^{\circ}$
Find $F$ so that $BE=BF=CE=DE, \angle{FBE}=108^{\circ}$, connect $AF, CF, DF, EF$. So

*

*$\angle{FBC}=108^{\circ}-36^{\circ}=72^{\circ}$


*$\triangle{BCE} \cong \triangle{BEF} \implies EF=BC=AB, \angle{BFE}=\angle{BEF}=36^{\circ}=\angle{BCE}=\angle{EBC} \\
\implies \angle{DEF}=36^{\circ}-24^{\circ}=12^{\circ} $


*$\angle{ABF}=108^{\circ}-78^{\circ}-18^{\circ}=12^{\circ}=\angle{DEF}$
$\implies \triangle{ABF} \cong \triangle{FED} \text{ [SAS]}  \\
\implies AF=DF, \angle{FAB}=\angle{DFE} \\
\implies \angle{FDA}=\angle{FAD} \\
\text{Also we have }\angle{BFE}=\angle{BCE} \implies BECF \text{ are cyclic} \\
\implies \angle{FCB}=\angle{BEF}=36^{\circ} \\
\implies \angle{BFC}=180^{\circ}-\angle{FCB}-\angle{FBC}=72^{\circ}=\angle{FBC} \\
\implies BC=FC=AC \\
\implies \text{C is circumcenter of } \triangle{ABF} \\
\implies \angle{AFB}=180^{\circ}-\dfrac{\angle{ACB}}{2}=180^{\circ}-30^{\circ}=150^{\circ} \\
\implies \angle{FAB}=\angle{DFE}=180^{\circ}-150^{\circ}-12^{\circ}=18^{\circ} \\
\implies \angle{BFD}=36^{\circ}-18^{\circ}=18^{\circ} \\
\implies \angle{DFA}=150^{\circ}-18^{\circ}=132^{\circ} \\
\implies \angle{FDA}=\angle{FAD}=\dfrac{180^{\circ}-132^{\circ}}{2}=24^{\circ} \\
\implies \angle{BAD}=\angle{FAD}-\angle{FAB}=24^{\circ}-18^{\circ}=6^{\circ} \\
\implies \angle{DAC}=\angle{BAC}-\angle{BAD}=60^{\circ}-6^{\circ}=\boxed{54^{\circ}}$
