Let $\triangle ABD$ be a triangle and $C$ be a point on $BD$ such that $ {AB} ={CD}$, $\angle BAC = 18^\circ$ and $\angle ABC = 12^\circ$. The question is to find $x$, where $\angle ADC = x^\circ$.
I think this is a pretty classic problem but I am not able to find a solution using the techniques which were used to solve a similar kind of problem (only the angles were different and they were $30^\circ$ and $24^\circ$ respectively). The idea was to create an equilateral triangle by drawing a perpendicular line segment from $B$ to $AC$ with the same length as $AB$. However, that technique does not work here and I cannot find another way. Any help is appreciated (I forgot to add this but an answer which uses "geometrical" methods like drawing line segments or forming new triangles etc. would be great), thank you. The figure is as follows:
Best Answer
While I was trying to correctly plot the angles in $\triangle ABC$ I figured out that an easy way is to use a regular triangle and a regular pentagon.
In the figure below you can see that, drawing the regular triangle $\triangle BCE$ on $BC$ and a regular pentagon on $BE$, the vertex of the pentagon is indeed in $A$. In fact it is known that $\angle EAB=36°$ and $\angle ABE=72°$ (you can easily verify it using symmetries in the pentagon), so $\angle ABC=\angle ABE-\angle CBE=72°-60°=12°$ and $\angle CAB=18°$ as it bisects $\angle EAB$.
When I see pentagons, what comes to my mind are golden triangles and golden ratios: $\triangle ABE$ is a golden triangle so $AB$ and $BC=BE$ are in golden ratio.
So we can build the golden triangle $\triangle KBC$ over $BC$ with $BK=CK=AB$. But the nice thing about golden triangles is that they "sum up": the triangle with $BD=BC+CD$ as a side and $BK$ as base is again a golden triangle as $\angle KBD=\angle KBC=72°$ and $BD$ and $BK$ are in golden ratio. So $DK=BD$ and $\angle BDK=36°$.
Now we can use again an equilateral triangle: $AB=BK$ and $\angle ABK=60°$ so $AK=AB$, the triangles $\triangle ABD$ and $\triangle ADK$ are equivalent as their sides have the same length, and finally $\angle BDA$ bisects $\angle BDK$.
So the solution is $\angle BDA=18°$.