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.
 A: I borrow (or rather steal!) an idea from @timon92's lovely answer to An angle inside a regular pentagon .
Draw a regular pentagon $BCKLM.$ Inside it, draw an equilateral triangle $BCO.$ From segments $OC, CK$ construct a rhombus $OCKA.$ Let the perpendicular from $K$ to $MB,$ which by symmetry bisects the angle $CKL,$ meet $CA$ at $D.$

As an easy (if inelegant) way to keep track of the various angles in the figure, define the "slope" of a directed line segment to be the angle that it makes with the directed line segment $MB.$
$BC$ has slope $72^\circ,$ therefore $OC$ has slope $12^\circ,$ therefore the parallel segment $AK$ also has slope $12^\circ.$
Angle $\angle KAD$ is half angle $\angle KAO,$ which is equal to $\angle OCK,$ which is $48^\circ$; therefore $\angle KAD = 24^\circ.$
Therefore $AD$ has slope $-12^\circ.$
But $DK$ has slope $90^\circ,$ therefore $\angle DKA = \angle KDA,$ therefore $AD = AK.$
Thus all the line segments shown in bold in the figure are the same length.
By congruent triangles, $\angle COD = \angle DKC = 54^\circ.$
But $\angle OBM = 48^\circ,$ and $\triangle OBM$ is isosceles, therefore $\angle MOB = \angle BMO = 66^\circ.$
Therefore:
$$
\angle COD + \angle BOC + \angle MOB = 54^\circ + 60^\circ + 66^\circ = 180^\circ.
$$
That is, the points $D, O,$ and $M$ are collinear.
(This is almost a straight copy of @timon92's argument regarding the point $G$ in the other question.)
By symmetry, $\triangle MBD$ is isosceles; and $\angle BMD = \angle BMO = 66^\circ$; therefore $\angle DBM = 66^\circ.$
We conclude that $\angle CBD = 108^\circ - 66^\circ = 42^\circ.$
But also $\angle DCB = 60^\circ + 24^\circ = 84^\circ.$
Therefore, $\triangle CBD$ in the above figure is similar to $\triangle CBD$ in the question.
Therefore, the figure $ABCD$ here is similar to the figure $ABCD$ in the question.
But $\angle BAC$ is the angle subtended by the chord $BC$ at the point $A$ of the circumference of the circle $ABC$ with centre $O.$ Because the angle at the centre is $\angle BOC = 60^\circ,$ it follows that $\angle BAC = 30^\circ.$
A: 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
$$\angle DAB = \frac 12 \angle DGB =\frac 12 60^\circ = 30^\circ.$$

A: 
Here is a mostly geometric solution.  First of all, let $C'$ be another point on $AC$ such that $BC'=BC$.  Observe that the triangle $BC'D$ has $\angle C'BD=30^\circ$ and $\angle BDC'=\angle BDC=54^\circ$.  Thus, by the Law of Sines on the triangle $BC'D$, we obtain
$$\frac{C'D}{C'B}=\frac{\sin(\angle C'BD)}{\sin(\angle BDC')}=\frac{\sin(30^\circ)}{\sin(54^\circ)}\,.$$
Since $\sin(54^\circ)=\cos(36^\circ)=\dfrac{1+\sqrt{5}}{4}$ (a proof is given here) and $\sin(30^\circ)=\dfrac{1}{2}$, we conclude that
$$\frac{C'D}{C'B}=\frac{-1+\sqrt{5}}{2}\,.$$
From $AD=BC=BC'$, we see that
$$\begin{align}\frac{C'A}{C'B}&=\frac{AD+C'D}{BC'}=1+\frac{CD'}{BC'}\\&=1+\frac{-1+\sqrt{5}}{2}
=\frac{1+\sqrt{5}}{2}=\frac{C'B}{C'D}\,.\end{align}$$
Hence, $C'A\cdot C'D=(C'B)^2$.  This shows that $C'B$ is a tangent to the circumcircle of the triangle $ADB$.  Therefore,
$$x=\angle BAC=\angle  BAC'=\angle C'BD=30^\circ\,.$$ 

Here is a proof that $\dfrac{C'D}{C'B}=\dfrac{-1+\sqrt{5}}{2}$ without using trigonometry (and thereby, proving that $\cos(36^\circ)=\dfrac{1+\sqrt{5}}{4}$).  Let $J$ be the reflection of $C'$ with respect to $BD$.  Therefore, $BC'J$ is an equilateral triangle.  If $r:=\dfrac{C'D}{C'B}$, then $r=\dfrac{C'D}{C'J}$.  

Draw a regular pentagon $C'JMLK$ so that $D$ is an interior point of this pentagon.  The thick line segments $JB$, $BC'$, $C'J$, $JM$, $ML$, $LK$, and $KC'$ are easily seen to be of the same length.  Note that $D$ is on the diagonal $C'M$ of the pentagon.  The triangle $DMJ$ is also an isosceles triangle with $DM=MJ$.  As $MJ=BC'$, we conclude that, in fact, $M=A$.
Thus, as $C'DJ$ and $C'JM$ are similar triangles,
$$r=\frac{C'D}{C'J}=\frac{C'J}{C'M}=\frac{C'J}{C'D+DM}\,.$$
(At this stage, since $A=M$, it already follows that $C'B^2=C'J^2=C'D\cdot CM=C'D\cdot C'A$, establishing that $C'B$ is a tangent to the circumcircle of the triangle $ADB$.  Hence, knowing the exact value of $r$ is unnecessary.)  Thus, from $DM=MJ=C'J$, we get
$$r=\frac{C'J}{C'D+C'J}=\frac{1}{\frac{C'D}{C'J}+1}=\frac{1}{r+1}\,.$$
That is, $r^2+r-1=0$, or $r=\dfrac{-1\pm\sqrt{5}}{2}$.  As $r>0$, we get $r=\dfrac{-1+\sqrt{5}}{2}$, as desired.
