Construct a pentagon $ABCDE$ if lengths of its sides are $a,b,c,d$ and $e$ ($|AB|=a,|BC|=b,...$), and the bisector of the angle at vertex $D$ is also the perpendicular bisector of $AB$.
I know that $D\in f$, $C\in c_1=c(B,b)$ and $E\in c_2=c(A,e)$.
How do I find the exact position of point $D$?
Let $x=AD=BD$. By the triangular inequality we have the constraints: $$ |b-c|<x<b+c \quad\text{and}\quad |d-e|<x<d+e. $$
For the perpendicular bisector of $AB$ to be also the bisector of $\angle CDE$ we must have $\angle ADE=\angle BDC$. By the cosine rule this implies: $$ {d^2+x^2-e^2\over 2dx}= {c^2+x^2-b^2\over 2cx}, $$ that is: $$ x^2={d(c^2-b^2)-c(d^2-e^2)\over c-d}. $$ Of course this gives a solution only if $x>a/2$ and if the above constraints are satisfied.
Example. Setting $$ a=4,\quad b=7,\quad c=4,\quad d=6,\quad e=4 $$ we get $x=\sqrt{139}$, which is larger than $b+c$. Hence no solution exists.
Let $E_{1}$ be the projection of $E$ onto $f$, and define $C_{1}$ similarly. Because $f$ bisects $\angle CDE$, $\triangle DC_{1}C\sim\triangle DE_{1}E$.
Let $\angle CDE = 2\theta$. Then, $DE_{1} = d\cos(\theta)$ and $DC_{1} = c\cos(\theta)$. Also, $EE_{1}=d\sin(\theta)$ and $CC_{1}=c\sin(\theta)$.
Let $A_{1}$ be the projection of $A$ onto $EE_{1}$ and let $B_{1}$ be the projection of $B$ onto $CC_{1}$. We have $EA_{1} = d\sin(\theta)-\frac{a}{2}$, so $AA_{1} = \sqrt{e^{2}-\left(d\sin(\theta)-\frac{a}{2}\right)^{2}}$ by the Pythagorean Theorem. Similarly, $BB_{1}=\sqrt{b^{2}-\left(c\sin(\theta)-\frac{a}{2}\right)^{2}}$.
Letting $M$ be the midpoint of $AB$, note that $ME_{1} = AA_{1}$ and $MC_{1} + BB_{1}$. However, we must have $ME_{1}+DE_{1} = MC_{1}+DC_{1}$. Thus:
$$d\cos(\theta)+\sqrt{e^{2}-\left(d\sin(\theta)-\frac{a}{2}\right)^{2}} = c\cos(\theta)+\sqrt{b^{2}-\left(c\sin(\theta)-\frac{a}{2}\right)^{2}}$$
This equation can be numerically solved for $\theta$ (possible analytically as well, although it may be too messy to be worth it). Then, $MD = d\cos(\theta)+\sqrt{e^{2}-\left(d\sin(\theta)-\frac{a}{2}\right)^{2}}$.
Defining $\alpha$ and $\beta$ as the exterior angles at $A$ and $B$, and $\theta$ as the complement of the half-angle at $D$, and $a_2:=a/2$ for convenience, "all we have to do" is solve the system
$$\begin{align} d \cos\theta &= a_2 + e \cos\alpha \tag1\\ c \cos\theta &= a_2 + b \cos\beta \tag2\\ d \sin\theta + e \sin\alpha &= c \sin\theta + b \sin\beta \tag3 \end{align}$$
The special case of $c=d$ clearly implies $b=e$ and $\alpha=\beta$, making the pentagon symmetric about the angle/perpendicular bisector, although its shape is not uniquely determined.
Otherwise ... A couple of rounds of squaring of $(3)$ will give us even powers of $\sin\alpha$ and $\sin\beta$, which we can replace with powers of $1-\cos^2\alpha$ and $1-\cos^2\beta$; substituting $\cos\alpha$ and $\cos\beta$ from $(1)$ and $(2)$, and simplifying, gives this quadratic in $\cos\theta$:
$$\begin{align} 0 &=\phantom{-}\;4\cos^2\theta\;(c - d) (d (b^2 - c^2) - c (e^2 - d^2)) \\ &\quad-4 \cos\theta\;a_2 (c - d) ((b^2 - c^2) - (e^2 - d^2))\\ &\quad-4 a_2^2 (c - d)^2 - (b + c - d - e) (b - c + d - e) (b + c - d + e) (b - c + d + e) \end{align}$$
Solving gives
$$\begin{align} \cos\theta &= \frac{a_2 (c - d) (b^2 - c^2 + d^2 - e^2)\pm \sigma}{2p (c - d)} \\[8pt] \sigma &:= \sqrt{ q(c - d) \left(a_2^2(c-d) + p\right)} \\[2pt] p &:= d(b^2-c^2)-c(e^2-d^2) \\[4pt] q &:= (b + c - d - e)(b + c - d + e)(b - c + d + e) (b - c + d - e) \end{align}$$
And then, from $(1)$ and $(2)$, $$\begin{align} \cos\alpha &= \frac{a_2(c-d)((c-d)(e^2-d^2)-p)\pm d\sigma}{2ep(c-d)} \\[4pt] \cos\beta &= \frac{a_2(c-d)((c-d)(b^2-c^2)-p)\pm c\sigma}{2bp(c-d)} \end{align}$$ where the $\pm$ represents the same sign throughout.
Of course, $\sigma$ must be real for a solution to exist. Whether one or the other (or both, or neither) states of the "$\pm$" leads to an extraneous solution is left as an exercise to the reader. $\square$
