Any ideas how I could find the size of $\angle CBD$ in the diagram given that AC=AD, $\angle CAB=6$, $\angle CBA=48$ and $\angle DAC=12$.
I think there should be a way to do it with basic geometry provided the right auxiliary lines are constructed.
The two things which led me to some progress where 1. Draw a perpendicular bisector from A to DC and an altitude from AB to C which gave me 3 congruent triangles, unfortunately I couldn't work and of the sides from CDB into it.
The second was to extend DC to make it the same length as AD and AC, this created another isosceles triangle with base angles of 48 degrees the base of this side made a cyclic quadrilateral with points A, B & C ... but unfortunately I still couldn't use the new information to solve for x (I was mainly looking to make congruent triangles)
We have: $\measuredangle ADC=84^{\circ}$ and $\measuredangle CDB=30^{\circ}-x.$
Thus, by Cheva we obtain: $$\frac{\sin(30^{\circ}-x)\sin48^{\circ}\sin12^{\circ}}{\sin{x}\sin6^{\circ}\sin84^{\circ}}=1$$ or
$$\frac{2\sin(30^{\circ}-x)\sin48^{\circ}}{\sin{x}}=1,$$ which gives $$x=18^{\circ}.$$
Indeed, since the exspression $\frac{\sin(30^{\circ}-x)}{\sin{x}}$ decreases on $(0^{\circ},30^{\circ})$,
there is an unique value of $x$ such that $$\frac{2\sin(30^{\circ}-x)\sin48^{\circ}}{\sin{x}}=1.$$ Id est, it's enough to prove that: $$\frac{2\sin(30^{\circ}-18^{\circ})\sin48^{\circ}}{\sin{18^{\circ}}}=1,$$ which is true because $$\frac{2\sin(30^{\circ}-18^{\circ})\sin48^{\circ}}{\sin{18^{\circ}}}=\frac{2\sin12^{\circ}\sin48^{\circ}}{\sin{18^{\circ}}}=\frac{\sin18^{\circ}-\sin18^{\circ}+\cos36^{\circ}-\cos60^{\circ}}{\sin{18^{\circ}}}=$$ $$=\frac{\sin18^{\circ}+\frac{2\sin36^{\circ}\cos36^{\circ}-2\sin36^{\circ}\sin18^{\circ}}{2\sin36^{\circ}}-\cos60^{\circ}}{\sin{18^{\circ}}}=$$ $$=\frac{\sin18^{\circ}+\frac{\sin72^{\circ}-\cos18^{\circ}+\cos54^{\circ}}{2\sin36^{\circ}}-\frac{1}{2}}{\sin{18^{\circ}}}=1.$$