How can I solve this geometry problem without trigonometry? 
Let $ABC$ be a triangle with $D$ on the side $AC$ such that $\angle DBC=42^\circ$ and $\angle DCB=84^\circ$.  If $AD = BC$, find $x = \angle DAB$.


It's supposed to be solved with constructions, but I couldn't figure out.  See the trigonometric version here.

Let $E$ be such that $BCDE$ is an isosceles trapezoid with bases $BC$, $DE$. Then 
$$\angle EBD = \angle EBC - \angle DBC = 84^\circ - 42^\circ = 42^\circ = \angle DBC = \angle BDE$$
thus triangle $EBD$ is isosceles with $BE=DE$.
Note that $ED=BE=DC$, $AD=BC$, and $\angle EDA = \angle BCD$. Hence triangle $EDA$ is congruent to $DCB$. Therefore, if we let $F$ be such that $DAFE$ is an isosceles trapezoid with bases $AD$, $EF$ then $ADEF$ is congruent to $BCDE$ and in particular $DF=DB$.
Let $G$ be a point such that triangle $BDG$ is equilateral. So, $B, G, F$ lie on a circle with center $D$. Hence $$\angle BFG = \frac 12 \angle BDG = 30^\circ.$$ Also, since $EF=EB$, we have $$\angle EFB = 90^\circ - \frac 12 \angle BEF = 90^\circ - 84^\circ = 6^\circ.$$
Thus $$\angle EFG=\angle EFB+\angle BFG=6^\circ+30^\circ=36^\circ.$$
On the other hand, clearly $\angle GED = 180^\circ -\frac 12\angle DEB = 180^\circ - \frac 12 96^\circ = 132^\circ$, hence 
$$\angle GEF = \angle GED - \angle FED = 132^\circ - 96^\circ = 36^\circ.$$
So $\angle EFG = \angle GEF$. It follows that $FG=EG$. Therefore $G$ lies on the perpendicular bisector of $EF$ which coincides with the perpendicular bisector of $AD$. Thus $GA=GD$.
Finally, since $GA=GD=GB$, points $A,D,B$ lie on a circle with center $G$. Hence
$$\angle DAB = \frac 12 \angle DGB =\frac 12 60^\circ = 30^\circ.$$

