Version 1.1

Quick Solution

Points $DABC$ are vertices of a regular pentagon.

$\angle ABC=108^\circ$ .

$\angle ADC=180^\circ-\angle ABC=\boxed{72^\circ}$

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The long version of the solution

Let's set $\angle ADC=a$

In $\triangle ADC$ we have $DC=AC\implies \angle DAC=\angle ADC=a$

Quadrilateral $ABCD$ is cyclic (from symmetry a perpendicular bisector of $BC$ intersects perpendicular bisector of $AD$ on the axis of symmetry, which is also the perpendicular bisector of $AB$ ).

Therefore $\angle ABC=180^\circ-\angle ADC=180^\circ-a$

In $\triangle ABC$ we have $AB=BC\implies \angle BAC=\angle BCA=\frac12(180^\circ-\angle ABC)=\frac12(180^\circ-(180^\circ-a))=\frac a2$

$\angle DAB=\angle DAC+\angle BAC=a+\frac a2=\frac {3a}2$

$\angle DAB=\angle ABC=180^\circ-a$

Combining the two lines above: $\frac{3a}2=180^\circ-a$

Solution of this equation is $a=\boxed{72^\circ}$

I agree with you, Marta, the problem is reduced to draw a regular pentagon of diagonal DB.

I might not be aware at first, but just a week ago I taught to my grandchildren how to do it, so it flash back to me .

I congratulate you for your geometrical puzzles so challenging and brain teaser... keep going