<s>[INST] 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.
 [/INST] 
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.
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