For reference: Given the quadrilateral ABCD such that $\angle CAD=25^\circ , \angle ACD=45^\circ$ and $\angle BAC=\angle BCA=20^\circ$ , what is the value of angle $\angle DBC$?
I made the drawing and distributed the angles.
I drew the perpendicular $CJ$ $\perp$ $JI$
I would need to demonstrate that $CJ$ is parallel to $BD$ and/or that $\angle CDB = \angle ICJ$ but I couldn't find the way
On
Here's an alternative short approach by slightly modifying the diagram you posted:
1.) Extend segment $BD$ to meet at point $E$ such that $AB=BC=BE$. Connect $E$ with $A$ and $C$. This gives $\triangle AEC$ and since $AB=BE=BC$, we can conclude that point $B$ is the circumcenter of $\triangle AEC$ and thus, via the inscribed angle theorem, $\angle AEC=\frac{\angle ABC}{2}=70$.
2.) We can see that $\angle AEC+\angle ADC=180$, therefore Quadrilateral $AECD$ is a cyclic quadrilateral. Using the properties of cyclic quadrilaterals, we can quickly establish that $\angle CED=\angle CAD=25$, this means that $\angle AED=45$. Since we know that $BC=BE$, this also means that $\angle CED=\angle BCE=25$, and thus, since $E$, $B$ and $D$ are collinear, $\angle DBC=25+25=50$.
In the figure, $E$ is a point on $CD$ produced such that $\angle DAE=20^o$
$ \because \angle CAE= \angle ECA=45^o$
$\therefore \angle AEC=90^o$ and $EA=EC$
Note that $BA=BC$, $\angle EAB=\angle ECB=65^o$ and $EA=EC$
implies that $\Delta EBA \cong \Delta EBC$ (SAS)
Hence $\gamma + \delta = \frac{140^o}{2}=70^o$ ----- (1)
Also $\alpha = \beta =45^o$
Since $\angle BAD =45^o= \alpha$
$\therefore A, B, D, E$ are concyclic.
Thus $\gamma = \angle DAE=20^o $ ----- (2)
(1) and (2) implies that $\delta =50^o$
Hence $\angle DBC=50^o$