Positioning Isosceles Triangle To Create Second Isosceles Triangle Two Triangles $ABC$ and $DEF$ are superimposed such that the points $D$ and $E$ fall on $AC$ and $BC$ respectively. $DEF$ is an isosceles triangle where $DF$ is equal to $EF$

Given the base and height of $DEF$ and the coordinates of points $(A_x,A_y)$, $(B_x,B_y)$, $(F_x,F_y)$:
find $(C_x,C_y)$ such that $DEC$ is also an isosceles triangle.
This seems like a relatively straight forward geometry problem, but I'm stumped. Any and all advice is welcome.
 A: Let $M=(M_x,M_y)$ be the midpoint of $DE$: point $M$ belongs to the circle of center $F$ and given radius $FH=h$, so that $(M_x-F_x)^2+(M_y-F_y)^2=h^2$. Once $M$ is chosen on that circle, then the coordinates of points $D$ and $E$ can be found, because they are the intersections between the line through $M$ perpendicular to $FM$ and the circle of center $F$ and given radius $FD=FE=l$.
You can then write the equations of lines $AD$ and $BE$, as a function of $M_x$ and $M_y$: their common point $C$ must lie on line $FM$, and that condition allows you to find $M$ and thus $C$.
A: Here's another approach.
We'll take $\triangle ABF$ to be the primary triangle; in what follows, $\angle A$ and $\angle B$, and $a$ and $b$, will indicate elements of that triangle. Let $A^\prime$, $B^\prime$, $D^\prime$, $E^\prime$ be feet of perpendiculars to the line through $F$ that is parallel to $\overline{AB}$. We'll take $\triangle DEF$ to be determined, not by its base and height, but equivalently, by its leg-length $r$ and vertex angle $2\phi$. Let the angle bisector of $\angle DFE$ meet $\overline{AB}$ at $G$, and introduce parameter $\theta = \angle AGF = \angle B^\prime F G$. 

Our goal is to determine $\theta$, which is established by the fact that $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, and $\overleftrightarrow{FG}$ meet at a common point, $C$. By the trigonometric form of Ceva's Theorem, this occurs if and only if   

$$\frac{\sin\angle BAD}{\sin\angle DAF} \cdot \frac{\sin\angle FBE}{\sin\angle EBA} \cdot \frac{\sin\angle AFG}{\sin\angle GFB} \;=\; 1 \tag{$\star$}$$

We have ...
$$\begin{align}
\angle AFG = \pi - \theta - \angle A &\quad\to\quad \sin\angle AFG = \sin(\theta + A) \\
\angle GFB = \phantom{\pi - \,}\theta - \angle B &\quad\to\quad \sin\angle GFB = \sin(\theta - B)
\end{align}$$
... and, by the Law of Sines, ...
$$\begin{align}
\frac{\sin \angle DAF}{r} = \frac{\sin \angle DFA}{d} &\quad\to\quad \sin\angle DAF = \frac{r}{d}\,\sin\angle DFA = \frac{r}{d}\,\sin(\theta+\phi+A) \\[6pt]
\frac{\sin\angle FBE}{r} = \frac{\sin\angle EFB}{e} &\quad\to\quad \sin\angle FBE = \frac{r}{e}\,\sin\angle EFB = \frac{r}{e}\,\sin(\theta-\phi-B)
\end{align}$$
... and also (finally utilizing all those perpendiculars)  ...
$$\begin{align}
\sin \angle BAD = \frac{|\overline{DD^{\prime\prime}}|}{d} = \frac{|\overline{AA^{\prime}}|-|\overline{DD^\prime}|}{d} = \frac{b \sin A - r \sin(\theta+\phi)}{d} \\[6pt]
\sin \angle EBA = \frac{|\overline{EE^{\prime\prime}}|}{e} = \frac{|\overline{BB^{\prime}}|-|\overline{EE^\prime}|}{e} = \frac{a \sin B - r \sin(\theta-\phi)}{e} 
\end{align}$$
With these, and a little cancellation, $(\star)$ becomes ...

$$\begin{align}
&\sin(\theta + A)\;\sin(\theta-\phi-B)\;\left(b \sin A - r \sin(\theta+\phi)\right) \\[4pt] =\; &\sin(\theta-B)\;\sin(\theta+\phi+A)\;\left(a \sin B - r \sin(\theta-\phi)\right)
\end{align} \tag{$\star\star$}$$

We an refine this slightly by introducing $p$ and $q$ as the circumdiameters of $\triangle ABF$ and $\triangle DEF$, respectively. The Law of Sines allows us to write ...
$$a = p \sin A \qquad b = p \sin B \qquad r = q \sin\angle FDE = q\cos\phi$$
Therefore, we have ...

$$\begin{align}
&\sin(\theta + A)\;\sin(\theta-\phi-B)\;\left(p \sin A \sin B - q \cos\phi \sin(\theta+\phi)\right) \\[4pt] =\; &\sin(\theta-B)\;\sin(\theta+\phi+A)\;\left(p \sin A \sin B - q \cos\phi \sin(\theta-\phi)\right)
\end{align} \tag{$\star\star\star$}$$

Solving this equation for $\theta$ is a considerable chore that I must postpone for now.
A: Another partial solution that I'm still working on.
As per the previous partial-solutions, we construct $M$ as the midpoint of $\overline{DE}$, so $C$ lies on $\overleftrightarrow{MF}$. We're given the "base and height" of $\triangle DEF$ and we assume the base is side $\overline{DE}$. Hence, we also know the height, $h$, which is the length of $\overline{FM}$.
The lines $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ respectively have equations:
$$y=\frac{C_y-A_y}{C_x-A_x}(x-A_x)+A_y\quad(1)$$
$$y=\frac{C_y-B_y}{C_x-B_x}(x-B_x)+B_y\quad(2)$$
Let $c_1$ be the circle with centre $F$ and radius $q=FD=FE$. Hence, the equation for $c_1$ is:
$$(x-F_x)^2+(y-F_y)^2=q^2\quad(3)$$
Let $c_2$ be the circle with centre $C$ and radius $r=CD=CE$. Hence, the equation for $c_2$ is:
$$(x-C_x)^2+(y-C_y)^2=r^2\quad(4)$$
Given that we know the length of $\overline{DE}$, we can say that $s=\overline{DE}$ and hence,
$$\sqrt{(D_x-E_x)^2+(D_y-E_y)^2}=s\quad(5)$$
As we are given $h=FM$, we may say that:
$$\sqrt{(M_x-F_x)^2+(M_y-F_y)^2}=h\quad(6)$$
Given that $c_1$ and $c_2$ intersect $\overleftrightarrow{AC}$ and $\overleftrightarrow{BC}$ at $D$ and $E$ respectively, we can say that $(D_x,D_y)$ and $(E_x,E_y)$ are the points that satisfy $(3)$, $(4)$ and $(5)$, with the additional criteria that $(D_x,D_y)$ satisfies $(1)$ and $(E_x,E_y)$ satisfies $(2)$.
Substituting $(1)$ into $(4)$ and using a computer a computer-algebra-system to solve for $x=D_x$ gives us that:
$$D_x=\frac{\pm r(A_x-C_x)\sqrt{(A_y-C_y)^2+(A_x-C_x)^2}+C_x(A_y-C_y)^2+C_x(A_x-C_x)^2}{(A_y-C_y)^2+(A_x-C_x)^2}$$
We can label the distance of $\overline{AC}$ as $AC$ and simplify this to:
$$D_x=\frac{\pm r(A_x-C_x)}{AC}+C_x$$
WLOG, we can say:
$$E_x=\frac{\pm r(B_x-C_x)}{BC}+C_x$$
We may also easily find each $D_y$, $E_y$ by substitution into $(1)$ and $(2)$.
At the moment, I'm getting a computer-algebra-system to combine the equations and will get back to you with any progress.
A: This is an attempt to try to find a solution by working backwards. That means starting with values for $A,B$, the lengths of $AC$ and $BC$ and $b$ and $h$ for the small triangle and trying to find $F$. The reason for doing it that way is that I can test my answers at each step along the way to catch mistakes.

So we know $(A_x, A_y)$, $(B_x, B_y)$, $L_a$, and $L_b$ which are the lengths of $AC$ and $BC$. We also know that $A_y = B_y$ which may simplify things a little. If we declare $A$ as the origin then we can say $AB$ is of length $B_x$
By the law of cosines, we can say 
$$\theta_{cab} = \arccos\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\tag{1}$$
We can then say that 
$$C_x = L_a\cos(\theta_{cab}) = L_a\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\tag{2}$$ 
and
$$C_y = L_a\sin(\theta_{cab}) = L_a\sin\left(\arccos\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)\right)$$
By the identity $\sin(arccos(x))= \sqrt{1-x^2}$ we simplify that to:
$$C_y = L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)^2}\tag{3}$$
We also know that $$\theta_{acb} = \arccos\left(\frac{L_a^2+L_b^2-B_x^2}{2L_aL_b}\right)\tag{4}$$
and that the lengths of $CD$ and $CE$ are given by
$$l_{CD}=l_{CE}=\pm\frac{b}{\sqrt2*\sqrt{1-\cos(\theta_{acb})}}=\pm\frac{b}{\sqrt2*\sqrt{1-\left(\frac{L_a^2+L_b^2-B_x^2}{2L_aL_b}\right)}}\tag{5}$$
We know this from simplifying the law of cosines for a equilateral triangle.
Where $$m_{AC} = \frac{A_y-C_y}{A_x-C_x} = \frac{A_y-L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}}{A_x-L_a\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}$$
Because $A_x = A_y = 0$ we get some cancellations and end up with
$$m_{AC} = \frac{L_a\sqrt{1-\left(\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}\right)}}{\frac{L_a^2+B_x^2-L_b^2}{2L_aB_x}}\tag8$$
I was going to try to keep everything as one equation and solve for $l_a$ and $l_b$ at the end but that's not looking good I'm going to stop writing every equation as a function of known variables now.
We know that
$$D_x = C_x - \frac{l_{CD}}{\sqrt{m_{AC}^2 + 1}}\tag6$$
$$D_y = m_{AC}*C_x\tag7$$
Why is this true? These equations are from old and messy notes. Where did I derive them from? $(7)$ comes from solving $m_{AC} = \frac{Ay-Dy}{Ax-Dx}=\frac{Dy}{Dx} \to D_y=m_{AC}D_x$
Similarly:
$$M_{BC} = frac{B_y-C_y}{B_x-C_x}=\frac{C_y}{C_x-B_x}\tag9$$
$$E_x = C_x - \frac{l_{CD}}{\sqrt{m_{BC}^2 + 1}}\tag{10}$$
$$E_y = m_{BC}*(E_x-B_x)\tag{11}$$
$M$ is the midpoint of $DE$ so:
$$M_x = \frac{D_x + E_x}{2}\tag{12}$$
$$M_y = \frac{D_y + E_y}{2}\tag{13}$$
The slope of the line $MF$ is given by $$m_{MF}=\frac{-1}{m_{DE}}=\frac{-1}{\frac{D_y-E_y}{D_x-E_x}} = \frac{E_x-D_x}{D_y-E_y}\tag{14}$$
We can now find $F_x$ and $F_y$ because we know the height of the triangle $DEF$, $M$ and the slope $m_MF$.
$$F_x = M_x - \frac{h}{\sqrt{m_{MF}^2 + 1}}\tag{15}$$
$$F_y = M_y - (M_x-F_x)M_{MF}\tag{16}$$
Now that we have good values to test against we can check our work along the way as we work in the other direction.
