Apollonius's Problem: why PCC can be reduced to PPC

Question

Here is the construction of the Apollonius's Problem PCC by deducing to PPC: https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/PCC.shtml But it is not mentioned how it works.

I'm trying to find the explanation, and I think it may be simplified as:

1. Draw two circles (A) and (C) tangent at B;
2. Draw a circle (E) tangent to (C) at D (here I make (A) and (E) intersected to avoid appearing 4 tangent lines in GeoGebra);
3. Draw two external tangent lines of (A) and (E) meeting at H, and F and G are two tangent points on one tangent line;
4. Draw a line passing through H and intersecting (C) at J and *K.

Prove that FGKJ are concyclic, which I haven't any idea yet.

Answer 1

I just find a proof by calculating angles of isosceles triangles BCD, BAF and DEG.

Denote ∠BAF as $2\alpha$ and DEG as $2\beta$, then ∠BCD is $2\alpha+2\beta$ because AF||EG, then $\angle BFG=\alpha$ and $\angle BDG=\angle BDC+\angle CDG=\pi-\alpha$, then BDGF are concyclic.

Because BDH are collinear (Monge & d'Alembert Three Circles Theorem II), we have: $$HB\cdot HD=HF\cdot HG=HJ\cdot HK$$ so FGKJ are concyclic.