This answer deals with the case where $b-a\lt c$ since lioness99a has already dealt with the case where $0\le c\le b-a$.
We may suppose that
$$A(0,a),\quad B(b,a),\quad C(b,0),\quad D(0,0),\quad E(c,0)$$
where $A,B,C,D$ are the vertices of the paper and we want to bend $B$ to $E$.
The equation of the perpendicular bisector $\ell$ of the line segment $BE$ is given by $$\ell : y-\frac a2=-\frac{b-c}{a}\left(x-\frac{b+c}{2}\right)$$
If we define $F,G,H,I$ as the intersection point of $\ell$ with $y=a,x=b,x=0,y=0$ respectively, we have, for $c\not=b$,
$$F\left(\frac{b^2-c^2-a^2}{2(b-c)},a\right),\ G\left(b,\frac{a^2-(b-c)^2}{2a}\right),\ H\left(0,\frac{a^2+b^2-c^2}{2a}\right),\ I\left(\frac{a^2+b^2-c^2}{2(b-c)},0\right)$$
We can see the followings :
If $c\lt b$, then $F_x\lt b$ and $0\lt G_y\lt a$ always hold and
$$F_x\ge 0\iff c\le \sqrt{b^2-a^2}$$
If $c\gt b$, then $F_x\ge b,G_y\lt a$ always holds and
$$G_y\gt 0\iff c\lt a+b\qquad\text{and}\qquad H_y\ge 0\iff c\le\sqrt{a^2+b^2}$$
Here, let us separate it into cases :
Case 1 : $b-a\lt c\le \sqrt{b^2-a^2}$
We see that $F$ is on the line segment $AB$ and that $G$ is on the line segment $BC$.
Therefore, the length of the bend is given by
$$FG=\sqrt{\left(\frac{b^2-c^2-a^2}{2(b-c)}-b\right)^2+\left(a-\frac{a^2-(b-c)^2}{2a}\right)^2}=\frac{(a^2+(b-c)^2)^{3/2}}{2a(b-c)}$$
Case 2 : $\sqrt{b^2-a^2}\lt c\le \sqrt{a^2+b^2}$
We see that $G$ is on the line segment $BC$ and that $H$ is on the line segment $AD$.
Therefore, the length of the bend is given by
$$GH=\sqrt{\left(b-0\right)^2+\left(\frac{a^2-(b-c)^2}{2a}-\frac{a^2+b^2-c^2}{2a}\right)^2}=\frac{b}{a}\sqrt{a^2+(b-c)^2}$$
Case 3 : $\sqrt{a^2+b^2}\lt c\lt a+b$
We see that $G$ is on the line segment $BC$ and that $I$ is on the line segment $CD$.
Therefore, the length of the bend is given by
$$IG=\sqrt{\left(b-\frac{a^2+b^2-c^2}{2(b-c)}\right)^2+\left(\frac{a^2-(b-c)^2}{2a}-0\right)^2}=\frac{a^2-(b-c)^2}{2a(c-b)}\sqrt{a^2+(b-c)^2}$$
Case 4 : If $c\ge a+b$, then we cannot bend the point.
The answer is as follows :
If $c\ge a+b$, then we cannot bend the point.
If $0\le c\lt a+b$, then the length of the bend is
$$\begin{cases}
\dfrac{a}{b-c}\sqrt{a^2+(b-c)^2} & \text{if $0\le c\le b-a$} \\\\
\dfrac{(a^2+(b-c)^2)^{3/2}}{2a(b-c)} & \text{if $b-a\lt c\le\sqrt{b^2-a^2}$} \\\\
\dfrac{b}{a}\sqrt{a^2+(b-c)^2} &\text{if $\sqrt{b^2-a^2}\lt c\le \sqrt{a^2+b^2}$}\\\\
\frac{a^2-(b-c)^2}{2a(c-b)}\sqrt{a^2+(b-c)^2} &\text{if $\sqrt{a^2+b^2}\lt c\lt a+b$} \end{cases}
$$
This doesn't produce a perfect pentagon. In the right triangle at the lower right corner of the rectangle, call the smaller angle $\alpha$ and the larger angle $\beta$. When you valley fold at $\heartsuit-\diamondsuit$ so that the stars coincide, this creates the angles indicated below.
Here's how it looks after the valley fold:
The next step (the mountain fold) hides the angle $\beta-\alpha$ behind the rest of the paper. The valley fold that follows will bisect the angle $\alpha + 45^\circ$. So when the origami is complete the folds you've made will create five angles in a half circle: one with measure $\beta-\alpha$ and four with measure $\frac12(\alpha + 45^\circ)$.
In a perfect pentagon, these angles would all be equal, i.e.,
$$
\beta - \alpha = \textstyle\frac12(\alpha + 45^\circ).
$$
Solving this, knowing that $\alpha + \beta=90^\circ$, this means in a regular pentagon we must have $\alpha=27^\circ$ and $\beta=63^\circ$. In particular this requires
$$
\tan\alpha=\tan(27^\circ) \approx 0.5095.$$
But for the advertised construction, we have $\tan\alpha = 0.5$. Close, but not exact!
Best Answer
It is all about continuity of the crumpling transform. The 1D case is easier to understand.
If you plot the abscissa of the points of the crumpled line vs. the abscissa of the corresponding point in the flat line, you get a continuous function in the range $[0,1]$. As the graph of the function splits the unit square horizontally and is continuous, it must meet the diagonal of the square.
(In other terms, the crumpled abscissa starts larger and ends smaller than the straight abscissa, so it must be equal somewhere.)
In the 2D case, you can plot both the crumpled abscissa and ordinate as a function of the flat abscissa and ordinate, which yield two surfaces in a unit cube. These surfaces meet the oblique planes $x=u$ and $y=v$ along two continuous curves, and these curves meet in at least one point.
This works even if the paper is elastic and can be stretched (i.e. topological matter).