Looking for a synthetic solution for this geometric problem

Question

The problem asks to find angle $\theta$. Its value is 42, so my goal is to prove that without using trigonometry.

Here is the following proof with trigonometry.

$AD/BD={{\sin42}\over {\sin \theta}}, AD/CD={{\sin 12}\over {\sin(96-\theta)}}, BD/CD={{\sin 6}\over {\sin 24}}$

$\implies$

$$1={{AD}\over {BD}}\cdot {{BD}\over {CD}}\cdot {{CD}\over {AD}}={{\sin 42}\over {\sin \theta}}\cdot {{\sin 6}\over {\sin 24}} \cdot {{\sin (96-\theta)}\over {\sin 12}}$$

Hence $\sin 96 \cot \theta-\cos 96={{\sin 24 \sin 12}\over {\sin 6 \sin 42}}$. Using the calculator we find that $\theta$ is 42 degrees.

For the synthetic solution I tried to extend BD until it meets AC in E. Now we have to prove that $\triangle ABD \sim \triangle BEA$. That's the only thing I saw that could be useful so far, then I don't have many more ideas (I tried more things that didn't lead to anything).

Answer 1

Let $E$ be the reflection of $B$ in $CD$. Since $\angle BDC = 180^\circ - \angle CBD - \angle DCB = 150^\circ$, it follows that $\angle CDE = 150^\circ$ and $\angle EDB = 60^\circ$. This shows that $BDE$ is equilateral.

Let $F\neq E$ be the second intersection of the circumcircle of $BDE$ and the line $CE$. Note that $\angle BFD = \angle BED = 60^\circ$, $\angle DFC = \angle DBE = 60^\circ$, and $\angle FCD = \angle DCB = 6^\circ$. Therefore $FD$ and $CD$ bisect angles of triangle $BCF$, hence $BD$ bisects the angle $CBF$. Thus $\angle DBF = \angle CBD = 24^\circ$.

Let $G$ be the intersection of $CD$ and $BF$. Note that $\angle GED = \angle DBG = 24^\circ$.

Let $H$ be the reflection of $G$ in $CE$. Note that $H$ lies on $DF$ because $\angle HFE = \angle EFG = 60^\circ = \angle DFC$. Also note that $\angle HCF = \angle FCG = 6^\circ$, hence $\angle HCD = 12^\circ$. Note that $\angle CEH = \angle GEC = \angle GED + \angle DEF = 24^\circ + \angle DBF = 48^\circ$. It follows that $\angle DEH = 48^\circ + 24^\circ = 72^\circ$. On the other hand, $\angle FDE = \angle FBE = \angle DBE - \angle DBG = 60^\circ - 24^\circ = 36^\circ$. It follows that $\angle EHD = 180^\circ - 72^\circ - 36^\circ = 72^\circ$. The triangle $DEH$ is therefore isosceles with $DH=DE$. So $DH=DB$, hence $\angle DBH = 90^\circ - \frac 12 \angle HDB = 90^\circ - \frac 12 \cdot 96^\circ = 42^\circ$.

We proved that $\angle HCD = 12^\circ$ and $\angle DBH = 42^\circ$. It follows that $H$ coincides with $A$. Since $DH=DB$, we have $\angle BHD = \angle DBH = 42^\circ$, as desired.