Two conics from six Thebault circles of a triangle [Significant edits, context, etc, provided by Blue, based on comments from the original author.]

While exploring Thebault circles in GeoGebra (see my interactive sketch), I noticed some interesting (and new?) connections with conics. I pose my conjectures as problems (1) and (2) below. 
I'm an electrical engineer, not a mathematician. I don't know how to prove these results.


Given concurrent cevians $\overleftrightarrow{AP}$, $\overleftrightarrow{BP}$, $\overleftrightarrow{CP}$ of $\triangle ABC$, consider the six Thebault circles (two per cevian) that these lines determine.
  
  
*
  
*Show that the centers of the Thebault circles lie on a conic.
  
*Each Thebault circle is, by definition, tangent to a side of the triangle (two circles per side). Show that the six points of tangency lie on a conic; further, show that the conic is a circle if $P$ is $\triangle ABC$'s orthocenter.



In the above, point $I$ is $\triangle ABC$'s incenter, which is known to be collinear with each pair of Thebault centers. Cut-the-Knot demonstrates a construction of a Thebault circle using the incenter.
[Images by Blue, too!]

Blue remarks: I've been able to verify (2) via this Ceva-like condition for conic-making ...
$$\frac{AF^\prime}{F^\prime B}\cdot
\frac{BD^\prime}{D^\prime C}\cdot
\frac{CE^\prime}{E^\prime A}\cdot
\frac{AF^{\prime\prime}}{F^{\prime\prime} B}\cdot
\frac{BD^{\prime\prime}}{D^{\prime\prime} C}\cdot
\frac{CE^{\prime\prime}}{E^{\prime\prime} A} = 1$$
... but my work is coordinate-based and very, very ugly; I'm hoping I'll find something better. Item (1) seems like it might follow reasonably from a Proposition that I stated in this answer, but I haven't attempted to work it through. What are the good proofs of these results?
 A: Here's a solution to the first part of (2):

The image as marked focusses attention on cevian $\overline{CF}$. Segments $\overline{IA}$, $\overline{IB}$, $\overline{FZ^\prime}$, $\overline{FZ^{\prime\prime}}$, $\overline{CJ^\prime}$, $\overline{CJ^{\prime\prime}}$ are angle bisectors, defining half-angles of size $\alpha$, $\beta$, $\phi^\prime$, $\phi^{\prime\prime}$, $\gamma^\prime$, and $\gamma^{\prime\prime}$. A little angle chasing shows that $\angle AIF^\prime \cong \angle ACJ^\prime$ and $\angle BIF^{\prime\prime} \cong \angle BCJ^{\prime\prime}$.
Circles $\bigcirc Z^\prime$ and $\bigcirc Z^{\prime\prime}$ are Thebault circles, constructed via the method shown at Cut-the-Knot.
By the Law of Sines in $\triangle AIF^\prime$ and $\triangle AIF^{\prime\prime}$, we have
$$\frac{|\overline{AF^\prime|}}{|\overline{IA}|} = \frac{\sin\gamma^\prime}{\sin\phi^{\prime\prime}} = \frac{\sin\gamma^\prime}{\cos\phi^\prime} \qquad\qquad
\frac{|\overline{AF^{\prime\prime}}|}{|\overline{IA}|} = \frac{\sin(\gamma^\prime + \pi/2)}{\sin\phi^{\prime}} = \frac{\cos\gamma^\prime}{\sin\phi^{\prime}}$$
Likewise, from $\triangle BIF^\prime$ and $\triangle BIF^{\prime\prime}$,
$$\frac{|\overline{F^\prime B|}}{|\overline{IB}|} = 
\frac{\cos\gamma^{\prime\prime}}{\cos\phi^{\prime}} \qquad\qquad
\frac{|\overline{F^{\prime\prime}B}|}{|\overline{IB}|} = \frac{\sin\gamma^{\prime\prime}}{\sin\phi^{\prime}}$$
Therefore,
$$\frac{|\overline{AF^\prime}|}{|\overline{F^\prime B}|} \cdot \frac{|\overline{AF^{\prime\prime}}|}{|\overline{F^{\prime\prime}B}|} =
\frac{|\overline{IA}|^2}{|\overline{IB}|^2}\;\frac{\sin\gamma^\prime}{\cos\gamma^{\prime\prime}}\cdot\frac{\cos\gamma^\prime}{\sin\gamma^{\prime\prime}} = \frac{|\overline{IA}|^2}{|\overline{IB}|^2}\;\frac{\sin 2\gamma^\prime}{\sin 2 \gamma^{\prime\prime}} = \frac{|\overline{IA}|^2}{|\overline{IB}|^2}\;\frac{\sin\angle ACF}{\sin\angle FCB}$$
Correspondingly, cevian $\overline{AD}$ would create Thebault circles with points of tangency $D^\prime$ and $D^{\prime\prime}$ on $\overline{BC}$; and cevian $\overline{BE}$ would create points of tangency $E^\prime$ and $E^{\prime\prime}$ on $\overline{CA}$ such that
$$\frac{|\overline{BD^\prime}|}{|\overline{D^\prime C}|} \cdot \frac{|\overline{BD^{\prime\prime}}|}{|\overline{D^{\prime\prime}C}|} = \frac{|\overline{IB}|^2}{|\overline{IC}|^2}\;\frac{\sin\angle BAD}{\sin\angle DAC} \qquad\qquad
\frac{|\overline{CE^\prime}|}{|\overline{E^\prime A}|} \cdot \frac{|\overline{CE^{\prime\prime}}|}{|\overline{E^{\prime\prime}A}|} = \frac{|\overline{IC}|^2}{|\overline{IA}|^2}\;\frac{\sin\angle CBE}{\sin\angle EBA}$$
which gives
$$\frac{|\overline{AF^\prime}|}{|\overline{F^\prime B}|} \cdot \frac{|\overline{AF^{\prime\prime}}|}{|\overline{F^{\prime\prime}B}|} \cdot
\frac{|\overline{BD^\prime}|}{|\overline{D^\prime C}|} \cdot \frac{|\overline{BD^{\prime\prime}}|}{|\overline{D^{\prime\prime}C}|} \cdot
\frac{|\overline{CE^\prime}|}{|\overline{E^\prime A}|} \cdot \frac{|\overline{CE^{\prime\prime}}|}{|\overline{E^{\prime\prime}A}|}
=
\frac{\sin\angle ACF}{\sin\angle FCB}\cdot
\frac{\sin\angle BAD}{\sin\angle DAC}\cdot
\frac{\sin\angle CBE}{\sin\angle EBA} $$
The right-hand side of this equation matches the trigonometric form of Ceva's Theorem; its value is $1$ if and only if cevians $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ concur at a point.
The left-hand side of the equation is a Ceva-like characterization for conics; its value is $1$ if and only if (distinct) points $D^\prime$, $D^{\prime\prime}$, $E^\prime$, $E^{\prime\prime}$, $F^\prime$, $F^{\prime\prime}$ lie on a conic. (See "Addendum" below.)
This proves the result. $\square$ 

For the second part of (2), consider the situation in which points $E^\prime$, $E^{\prime\prime}$, $F^\prime$, $F^{\prime\prime}$ lie on a circle.

By the Power of a Point theorem, we can calculate the power of $A$ with respect to the circle in two ways to get equivalent expressions:
$$\operatorname{pow}(A) = |\overline{AF^\prime}||\overline{AF^{\prime\prime}}| = |\overline{AE^{\prime\prime}}||\overline{AE^\prime}|$$
(This condition happens to be necessary and sufficient.) From our previous work, we can re-write that equality as
$$\begin{align}|\overline{IA}|^2\;\frac{\sin\gamma^\prime}{\cos\phi^\prime}\;\frac{\cos\gamma^\prime}{\sin\phi^\prime} = 
|\overline{IA}|^2\;\frac{\sin\beta^{\prime\prime}}{\cos\epsilon^\prime}\;\frac{\cos\beta^{\prime\prime}}{\sin\epsilon^\prime} &\quad\to\quad
\frac{\sin 2\gamma^\prime}{\sin 2\phi^\prime} = 
\frac{\sin 2\beta^{\prime\prime}}{\sin 2\epsilon^\prime} \\[4pt]
&\quad\to\quad
\frac{\sin \angle ACF}{\sin \angle AFC} = 
\frac{\sin \angle ABE}{\sin \angle AEB} \\[4pt]
&\quad\to\quad
\frac{\sin(A+\angle AFC)}{\sin \angle AFC} = 
\frac{\sin(A+\angle AEB)}{\sin \angle AEB} \\[4pt]
\end{align}$$
Certainly, if cevians $\overline{CF}$ and $\overline{BE}$ are altitudes, then this relation is satisfied (since $\angle AFC$ and $\angle AEB$ would be congruent (right) angles). Further, concurrent cevian $\overline{AD}$ would also be an altitude, so that the corresponding relationships involving $D^\prime$ and $D^{\prime\prime}$ would also be satisfied, so that all six Thebault tangency points would like on a common circle. (The reader may consider whether it's possible for non-altitudes to satisfy the relations.) This concludes the second part of $(2)$. $\square$

Addendum. I wasn't aware of the Ceva-like characterization for conics until I worked on this solution. For proof, one can use coordinates to define, say, 
$$A := (x_A, y_A) \qquad B := (x_B, y_B) \qquad C := ( x_C, y_C )$$
$$D^\prime := \frac{ B + C\;d^\prime}{1 + d^\prime} \qquad 
E^\prime := \frac{ C + A\;e^\prime}{1 + e^\prime} \qquad
F^\prime := \frac{ A + B\;f^\prime}{1 + f^\prime}$$
$$D^{\prime\prime} := \frac{ B + C\;d^{\prime\prime}}{1 + d^{\prime\prime}} \qquad 
E^{\prime\prime} := \frac{ C + A\;e^{\prime\prime}}{1 + e^{\prime\prime}} \qquad
F^{\prime\prime} := \frac{ A + B\;f^{\prime\prime}}{1 + f^{\prime\prime}}$$
Writing $P_x$ and $P_y$ for the $x$- and $y$-coordinates of point $P$, the last six points lie on a conic if and only if they satisfy this relation (see Equation (8) in MathWorld's "Conic Section" entry):
$$\left|\begin{array}{ccccccc}
(D^\prime_x)^2 & D^\prime_x D^\prime_y & (D^\prime_y)^2 & D^\prime_x & D^\prime_y & 1 \\
(E^\prime_x)^2 & E^\prime_x E^\prime_y & (E^\prime_y)^2 & E^\prime_x & E^\prime_y & 1 \\
(F^\prime_x)^2 & F^\prime_x F^\prime_y & (F^\prime_y)^2 & F^\prime_x & F^\prime_y & 1 \\
(D^{\prime\prime}_x)^2 & D^{\prime\prime}_x D^{\prime\prime}_y & (D^{\prime\prime}_y)^2 & D^{\prime\prime}_x & D^{\prime\prime}_y & 1 \\
(E^{\prime\prime}_x)^2 & E^{\prime\prime}_x E^{\prime\prime}_y & (E^{\prime\prime}_y)^2 & E^{\prime\prime}_x & E^{\prime\prime}_y & 1 \\
(F^{\prime\prime}_x)^2 & F^{\prime\prime}_x F^{\prime\prime}_y & (F^{\prime\prime}_y)^2 & F^{\prime\prime}_x & F^{\prime\prime}_y & 1 \\
\end{array}\right| = 0$$
Expanding and factoring the determinant gives
$$\frac{|\triangle ABC|^4\;(d^\prime-d^{\prime\prime})(e^\prime-e^{\prime\prime})(f^\prime-f^{\prime\prime})(d^\prime d^{\prime\prime} e^\prime e^{\prime\prime} f^\prime f^{\prime\prime}-1)}{(1+d^\prime)(1+d^{\prime\prime})(1+e^\prime)(1+e^{\prime\prime})(1+f^\prime)(1+f^{\prime\prime})} = 0$$
so that (ignoring degeneracies, and taking the six points to be distinct) we have this condition:

$$d^\prime d^{\prime\prime} e^\prime e^{\prime\prime} f^\prime f^{\prime\prime}\;=\;1$$

To get the form used in the solution above, we simply observe that $d^\prime = \frac{|\overline{BD^\prime}|}{|\overline{D^\prime C}|}$, etc. $\square$
