Solve the angle $\angle{DCB}$ in triangle $\triangle{ABC}$ with $\angle{A}=84^{\circ}$

Question

Where $\angle{A}=84^{\circ}, \angle{ACD}=42^{\circ}, BD=AC$, find $\angle{BCD}$.

Wonder if there is solution without using trigonometric functions.

I tried with getting circumcenter of triangle ABC, but seems hard to prove it forms an equilateral triangle with side AC.

Also if trying from equilateral and form an Isosceles triangle with two angles of $24^{\circ}$, and then form another isosceles triangle with top angle to be $\angle{B}=24^{\circ}$, it is not easy to prove that $\angle{ADC}=54^{\circ}$.

Answer 1

Draw isosceles triangle $ACE$ such that $CA=CE$.

Let the circumcentre of $\triangle DEC$ be $F$ and draw the equilateral triangle $DEF$.

$\angle CEF=\angle ECF=36^\circ$.

Now construct another equilateral triangle with side $CE$ as in the figure.

$\angle DEG=36^\circ$

$\therefore\triangle DEG\cong\triangle FEC\text{ (S-A-S)}\\ \implies DE=DG.$

Also $\angle BDG=72^\circ.$

Then mark point $H$ on $CE$ such that $CF=CH$.

$\therefore \angle CHF=\angle CFH=72^\circ$.

Also $\triangle EHF$ is isosceles.

Now we can see $\triangle EDG\sim\triangle EHF$. From there we can prove that $\triangle BGE\sim \triangle CFE$.

From similar isosceles triangles above, $\frac{ED}{EG}=\frac{EH}{EF}.$
Adding $1$ to both sides, $\frac{EB}{EG}=\frac{EC}{EF}\implies\small \triangle BGE\sim \triangle CFE.$

It shows that $BG=GE$ and then we can easily show $\angle EBC=\angle ABC=30^\circ.$