When do circles on the sides of a triangle intersect? 
If you have a non degenerate triangle and two of the sides are chords of two respective circles, then under what conditions do the circles intersect at two distinct points?

I'm having trouble understanding why this condition holds in a step of a proof of a theorem.
Here's the statement and proof of the theorem:
Source: http://www.cut-the-knot.org/proofs/nap_circles.shtml

Theorem
Let triangles be erected externally on the sides of ΔABC so that the sum of the "remote" angles P,Q, and R is 180°. Then the circumcircles of the three triangles ABR, BCP, and ACQ have a common point.
Proof
Let F be the second point of intersection of the circumcircles of ΔACQ and ΔBCP. ∠BFC + ∠P = 180°. Also, ∠AFC + ∠Q = 180°. Combining these with ∠P + ∠Q + ∠R = 180° immediately yields ∠AFB + ∠R = 180°. So that F also lies on the circumcircle of ΔABR.

The bolded line is what my question is about. Why is this true in this case, and when is it true in general?
I've thought of a case where this would not be true: if you have $3$ circles tangent to each other and take the triangle formed by connecting the 3 points of contact.


Let we say that $\Gamma_C$ is a circle through $A,B$ and $\Gamma_B$ is a circle through $A,C$. Since $\Gamma_B$ and $\Gamma_C$ meet at $A$, they have for sure a second intersection point, unless they are tangent at $A$. That implies that their centres $O_B,O_C$ and $A$ are collinear, so if $P,Q$ are the remote points on $\Gamma_B$ and $\Gamma_C$ respectively,
$$\widehat{APC}+\widehat{AQB}=\frac{\widehat{AO_B C}+\widehat{A O_C B}}{2}=\pi-\left(\widehat{O_B A C}+\widehat{O_C A B}\right)=\widehat{BAC}.$$
That implies that the third remote point $R$ has to fulfill $\widehat{BRC}=\pi-\widehat{BAC}$, so the circumcircle of $BCR$ goes through $A$ and the theorem is still true in this slightly degenerate case.
