The Twin

Using the formula $r * s = [\triangle ADB]$ where $s$ is the semiperimeter, $r$ is the radius of the triangle's incircle and $[\triangle ADB]$ is the area of $\triangle ADB$ , we can solve for the radius. Because $\overline {\rm BD} = 5$ by the Pythagorean Theorem, $s = \frac {3 + 4+ 5}{2} = 6$ . Furthermore, $[\triangle ADB] = \frac{3*4}{2} = 6$ . Thus, solving for the radius of either incircle, we get $r = \frac{ [\triangle ADB]}{s} = \frac{6}{6} = 1$ Then, by realizing that both triangles are identical, $\overline {\rm DE}$ must be $1$ apart vertically and $2$ apart horizontally. Thus, by the Pythagorean Theorem. $\sqrt{1^{2} + 2^{2}} = \sqrt{5} \approx \boxed{2.236}$

Consider my diagram.

$\tan \angle DBC=\dfrac{3}{4}$ $\implies$ $\angle DBC = \tan^{-1}\left(\dfrac{3}{4}\right)$

It follows that

$\angle EBC = \dfrac{1}{2} \angle DBC = \dfrac{1}{2}\left[\tan^{-1}\left(\dfrac{3}{4}\right)\right]$

Then,

$\tan \angle EBC = \dfrac{r}{4-r}$

$\tan \dfrac{1}{2}\left[\tan^{-1}\left(\dfrac{3}{4}\right)\right]=\dfrac{r}{4-r}$

By the use of a calculator,

$\dfrac{1}{3}=\dfrac{r}{4-r}$

$r=1$

Apply pythagorean theorem on the blue right triangle, we have

$(DE)^2=(3-2r)^2+(4-2r)^2=(3-2)^2+(4-2)^2=1+4=5$

Finally,

$\boxed{DE=\sqrt{5}\approx 2.236067978}$

Circles $D$ and $E$ are incenters of $\triangle ABD$ and $\triangle CDB$ . Setting $B$ at the origin, we have $A(0, 3)$ , $B(0, 0)$ , $C(4, 0)$ , and $D(4, 3)$ .

So $\triangle ABD$ 's incenter $D$ is at $(\frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c}) = (\frac{3 \cdot 4 + 4 \cdot 0 + 5 \cdot 0}{3 + 4 + 5}, \frac{3 \cdot 3 + 4 \cdot 0 + 5 \cdot 3}{3 + 4 + 5}) = (1, 2)$ , and $\triangle CDB$ 's incenter $E$ is at $(\frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c}) = (\frac{3 \cdot 0 + 4 \cdot 4 + 5 \cdot 4}{3 + 4 + 5}, \frac{3 \cdot 0 + 4 \cdot 3 + 5 \cdot 0}{3 + 4 + 5}) = (3, 1)$ .

The distance between $D(1, 2)$ and $E(3, 1)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{5} \approx \boxed{2.236}$ .

Radius of the circles is 1. Horizontal distance between centers is 2, vertical distance is 1. $DE=\boxed{\sqrt5}$ .