Folding a corner on a paper to the opposite side If we bend the $\text{A4}$ paper by connecting two opposite corners, what length is the $\color{red}{\text{bend}}$?

I drew the two blue lines on the last image, which are sides of a rhombus. By the help of that rhombus, I managed to calculate the bend as the length of the shorter diagonal: 
$$\frac ab\sqrt{a^2+b^2}$$
Taking the length of paper sides as $1$ and $\sqrt2$, the bend is then $\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}$.


But what if we bend the point not to the opposite corner, but to the
  some point $T$ on the oppoiste side of the rectangle with sides $a,b$ where that point is $c$ units away from the bottom left corner?

When $c=0$, we get $B=\frac ab\sqrt{a^2+b^2}$ as the above case.
 When $c=b$, we get $B=b$, as you are just folding the paper then.
But how do we solve for the $B$ in general over some $a,b,c$? 
We can also extend the line of the bottom side and put our point $T$ outside of the paper, as long as the bottom left point is in the same place; and consider these cases too.
 A: While $0\leq c\leq b-a$, we can just consider this as a piece of paper which has size $a$ by $b-c$:

Therefore, the equation to calculate the length of the bend will be \begin{align}B&=\frac a{b-c}\sqrt{a^2+(b-c)^2}\end{align}
When $c=b-a$ then the bend starts in the bottom right corner of the paper, and so this approach won't work for $c> b-a$ as it no longer forms a rhombus.
[I will update my answer with the other half when I've had time to do the Maths]
A: 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 
$$\begin{cases}
\dfrac{a}{b-c}\sqrt{a^2+(b-c)^2}  & \text{if $0\le c\le b-a$} \\\\
\dfrac{(a^2+(b-c)^2)^{3/2}}{2a(b-c)} & \text{if $b-a\lt c\le\sqrt{b^2-a^2}$} \\\\
\dfrac{b}{a}\sqrt{a^2+(b-c)^2} &\text{if $\sqrt{b^2-a^2}\lt c\le \sqrt{a^2+b^2}$}\\\\
\frac{a^2-(b-c)^2}{2a(c-b)}\sqrt{a^2+(b-c)^2} &\text{if $\sqrt{a^2+b^2}\lt c\lt a+b$} \end{cases}
$$
