Let D be point in the interior of ABC, Given that Angle DAC = $30°$ and Angle DCA = $40°$ , AB = BD + DC, BD = AC. Find the measure of DBC Let D be a point inside triangle ABC.
The following are given:
Angle DAB = $30°$
and Angle DBA = $40°$
, AC = CD + DB, CD = AB
Find angle DCB.
I tried extending BD to D' such that ABD' and DD'C would both be isosceles, but I cannot find anything after that. I don't know how to utilize AC = BD,  I tried cutting and pasting triangles in the figure but found no success. I also need a synthetic solution. I also tried extending B to G as shown in the figure, but I can't prove any of the statements


Your observations are accurate and to prove some of them without trigonometry, here is how I went about.

$\angle ADC$ is obtuse ($110^0$) and hence the circumcenter of $ \triangle ADC$ will be outside the triangle and closer to line $AC$. As per cyclic quadrilateral,
$\angle AFC = 180^0 - \angle ADC = 70^0$
Hence, $\angle AOC = 140^0, \angle OAC = \angle OCA = 20^0$.
As $\angle ACD = 40^0, \angle OCD = 60^0$; so, $\angle ODC = 60^0$.
So, $ \triangle OCD$ is equilateral triangle and $CD = OC \,$ which is radius of the circumcircle of triangle $ACD$.
Now extend line $AC$ to point $E$ such that $CE = CD = OC$.
$\angle DCE = 140^0 \,$ so, $\angle CDE = \angle DEC = 20^0$.
$\triangle OAC \cong \triangle CDE \,$ by side-angle-side. So, $DE = AC$.
$AE = AB = AC + CD$
$\triangle ADE \cong \triangle ADB$ by side-side-side. So, $\angle DAB = 30^0, \angle DBA = 20^0$.
$\angle BDE = 100^0$ and $BD = DE. \,$ So, $\angle DBE = \angle DEB = 40^0$.
As $\, \angle DCE + \angle DBE = 140^0 + 40^0 = 180^0$, $BDCE$ is a cyclic quadrilateral.
Since chord $CD = CE$, $\angle DBC = \angle CBE = 20^0$.
That also proves $\angle DCB = 40^0$ and that point $D$ is incenter of $\triangle ABC$.
