Unable to find similarity of two triangles.

Question

In the image below taken from article title "Trisection using Bisection", here; the issue lies in proving that $ DH/DA = DH/DJ = DJ/DB = DA/DB = \frac{1}{3}$.

How the link has been taken to $DA$ is not understandable. Request how to get to the geometrical proof that infers from the triangle similarity below that the ratio $DH/DA =\frac{1}{3}$. It seems that there is help taken of algebra to find the shifted center of the circle $c2$, and the simple fact that $AD = \frac{AB}{4}$.

There is stated to be similarity between the triangle $DJB, DHJ$, which can have one of three options ($AAA, SSS, SAS$).

Can only state that one angle is similar (i.e. $90^0$). Could find nothing else.

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Edit Use is made of two facts:
(i) $AD=DC, AC=CB\implies AD = DB/3$;
(ii)$DJ=DA$, as both radius of circle $c1$.
As the similarity shows $DH/DJ=DJ/DB$, so the rest of the argument can be constructed easily.

Answer 1

The triangles are similar because their angles are the same (AAA). They both have an angle of $90^\circ$, and share an angle at $D$ (so of course the third angle is shared as well).

Incidentally, the labelling of the vertices in your diagram is wrong: the second triangle should be $DHJ$, not $BHJ$.

Answer 2

The trick is to realize that $\angle DJB = 90^\circ$ because it is subscribed in a a semi-circle (https://www.mathopenref.com/semiinscribed.html)

So $\angle DJH \cong \angle JBH$ and $\angle JDH \cong \angle BJH$