My version is not a strict formal proof, but rather an intuition behind the idea. I hope it helps to get the point without having to dive into a lot of formal steps and theorems.

There are actually two major symmetries in this construct that make the whole thing possible.

The first is the reflection symmetry of the circle *p* relative to the *AB* line. It guarantees that the two arcs, *DB* and *BE*, are equal, which in turn implies that *DE* arc is twice as long as *DB* arc.

Now pretend this construct in motion. When we move point *B* along the circle *p*, keeping all given constraints, the *DE* segment rotates relative to the *AD* line twice as fast as *DB* does and, consequently, $\angle ADE$ changes twice as fast as $\angle ADB$.

Here we have to note the second major symmetry -- the symmetry of the circles *q* and *s* relative to the *DB* line. This symmetry implies that $\angle EDB = \angle GDB$.

Continuing to look at this in motion, we notice that when we move *B*, say, towards *D*, the *DE* segment 'chases' the *DB* line with twice its speed. At the same time, the *DG* segment symmetrically 'meets' the *DB* line with the same speed as *DE*, but in the **opposite** direction. This neatly guides us to the notion that the *DG* segment stays fixed, or, in other words, $\angle ADG = const$.

Now all we're left to do is to find the actual value of $\angle ADG$. We can continue to look at the motion described above, pretending that *B* approaches *D*. We see that all the lines *DE*, *DB*, and *DG* approach each other and become one line as *B* becomes *D*. And this line is clearly tangent to the circle *p*. But *DG* was not even moving relative to *AD*, so it should have been the same tangent line from the very beginning.

Hope that helps.