Given $\triangle ABC$, $BC=a,\ AC=b,\ AB=c$;
$\angle BAD=\angle CAD$,
$BE\perp AC$, $AF=BF$,
$\angle BAC=\alpha$,
$\angle BCA=\gamma$.
First, define coordinates of the points $A,B,C,E$ and $F$:
\begin{align}
A &=(0,0),\\
B &=(c\cos\alpha,c\sin\alpha), \\
C &=(b,0),\\
E &=(c\cos(\alpha),0),\\
F &=(c/2\cos\alpha,c/2\sin\alpha) \\
\end{align}
Coordinates of the point $D$ can be defined as follows:
\begin{align}
D_y &= x\tan\gamma = (b-x)\tan\frac{\alpha}{2},
\\
x&=\frac{b}{\tan\gamma+\tan\frac{\alpha}{2}},
\\
D&=\left(
\frac{b}{\tan\frac{\alpha}{2}(1+\tan\frac{\alpha}{2}/\tan\gamma)}
,\frac{b}{1+\tan\frac{\alpha}{2}/\tan\gamma}
\right).
\end{align}
Next, find the intersection points
$U=AD \cap CF$,
$V=AD \cap BE$,
$W=BE \cap CF$:
\begin{align}
U&=\left(
c b \frac{1+\cos\alpha}{c+2 b}
,
c b \frac{\sin\alpha}{c+2 b}
\right)
\\
V &= \left(
c\cos\alpha, c\cos\alpha\frac{(1-\cos\alpha)}{\sin\alpha}
\right)
\\
W &= \left(c \cos(\alpha),c\sin(\alpha)
\frac{b-c\cos(\alpha)}{2b-c\cos(\alpha)}
\right)
\end{align}
\begin{align}
S_{\triangle UVW}
&=
|(V_x-U_x)(W_y-U_y)-(V_y-U_y)(W_x-U_x)|/2.
\end{align}
Finally, after substitution $\sin\alpha=\sqrt{1-\cos^2\alpha}$,
$\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$
and simplification,
the area of $\triangle UVW$ is given
in terms of $a,b$ and $c$
by quite a nice construction:
\begin{align}
S_{\triangle UVW}
&=
\frac{1}{4}
\sqrt{\frac{(a-b+c)(a+b-c)}{(a+b+c)(b+c-a)}}
\frac{\left(b^3-b^2 c+b c^2+c^3-a^2 (b+c)\right)^2}
{b|3 b^2-c^2+a^2| (c+2 b)}
.\quad(*)
\end{align}
Test:
the first triangle: $a=8$, $b=10$, $c=12$,
$S_{\triangle UVW}=3.266386$,
the second triangle: $a=8$, $b=10$, $c=15$,
$S_{\triangle UVW}=16.835510$.
And as expected, when $a=b=c$, $S_{\triangle UVW}=0.$
Edit:
Equation (*) also reveals another extreme case,
when the altitude $BE$ is parallel to the median $CF$ and
$S_{\triangle UVW}=\infty$ due to the condition $3 b^2-c^2+a^2=0$
in the denominator:
Indeed in this case, $c^2-(2b)^2=a^2-b^2$.
Best Answer
Let $x$ be the common $A$-bisector/$B$-median/$C$-altitude.
In its role as the $A$-bisector, $x$ satisfies (by Stewart's Theorem and the Angle Bisector Theorem) $$b^2\,\frac{ac}{b+c}+c^2\,\frac{ab}{b+c}=a\left(x^2+\frac{a^2bc}{(b+c)^2}\right)\;\to\;x^2(b+c)^2=bc(a+b+c)(-a+b+c) \tag{1}$$
As the $B$-median (again by Stewart, or Apollonius), $$c^2\;\frac{b}{2}+a^2\;\frac{b}{2} = b \left(x^2+\frac{b^2}{4}\right)\quad\to\quad 4x^2 = 2a^2-b^2+2c^2 \tag{2}$$
As the $C$-altitude (invoking Heron's formula), $$\frac12cx=|\triangle ABC|\quad\to\quad4c^2x^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{3}$$
The non-linear system $(1)$, $(2)$, $(3)$ turns out to be quartic. Throwing it into Mathematica for numerical solution, and discarding non-positive values, gives two options:
(The values in my comment to the question correspond to taking $c=1$.) The first solution is, of course, the equilateral triangle; the second is obtuse:
So, if we restrict ourselves to acute triangles, the equilateral becomes the only solution. $\square$