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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.

Three circles

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  • $\begingroup$ In the last configuration, the sum of the angles made by remote points is $\frac{\pi}{2}$, not $\pi$. $\endgroup$ Sep 1, 2015 at 14:36

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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.

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… under what conditions do the circles intersect at two distinct points?

When the two poionts of intersection don't coincide, i.e. if the circles don't have a common tangent. So when do they touch? Two circles which are known to have one point in common are touching iff their centers are collinear with the common point.

Let $F$ be the second point of intersection of the circumcircles of $\triangle ACQ$ and $\triangle BCP$.

This doesn't neccessarily imply distinct: $F$ might well be eqial to one of $A,B,C$. The answer by Jack D'Aurizio covered this degenerate situation in some detail.

… if you have 3 circles tangent to each other and take the triangle formed by connecting the 3 points of contact.

For simplicity take the regular case you depicted, where all three circles have the same size. Then the chords would form a regular triangle with each of the circle centers, which means that another point on the outer arc would form a $30°$ angle with each chord, so the angle sum would be $90°$, in violation to the requirements.

Rebuttal of counterexample

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