Triangle inside triangle There is a triangle with $A$, $B$ and $C$ points. And there is another small triangle inside with $D$, $E$ and $F$ points. All of three sides of each triangle are parallel and all of the distances between these sides are equal: $|GH| = |IJ| = |KL| = k$
I have $A$, $B$ and $C$ points $(x_a,y_a)$,
 $(x_b,y_b)$, $(x_c,y_c)$ and distance between sides $k$. How can I find the smaller triangle's points $(D, E, F)$.

 A: If you can find the coordinates of $D$ relative to $A$ you can find the other coordinates by the same method.
Write $\overrightarrow{AB}=\underline{a}$ and $\overrightarrow {AC} = \underline{b}$ and let $\overrightarrow{AD}=\underline{x}$
Then since D lies on the angle bisector of angle $BAC$, we have $$\underline{x}=\lambda(\underline{\hat{a}}+\underline{\hat{b}})$$
Here the hats denote the unit vectors.
Let angle $BAD=\theta$ and $\underline{\hat{n}}$ be the unit vector normal to the plane. Then, using cross-product,$$\underline{x}\times\underline{a}=|\underline{x}||\underline{a}|\sin\theta\underline{\hat{n}}=k|\underline{a}|\underline{\hat{n}}$$
$$\implies\lambda(\underline{\hat{a}}+\underline{\hat{b}})\times\underline{a}=k|\underline{a}|\underline{\hat{n}}$$
$$\implies\lambda=\frac{k}{|\underline{\hat{b}}\times\underline{\hat{a}}|}=\frac{k|\underline{a}||\underline{b}|}{2\Delta}$$
Here $\Delta$ denotes the area of the triangle $ABC$
Thus the position of $D$ relative to $A$ is given by $$\underline{x}=\frac{k}{2\Delta}\left(\underline{a}|\underline{b}|+\underline{b}|\underline{a}|\right)$$
This can be written easily in terms of the given coordinates.
A: $D$ is at equal distance to $AB$, $AC$, so it lies on the angle bisector from $A$. In the same manner, $E,F$ are on the angle bisectors from $B,C$ respectively.
Consider $I$, the incenter, the point of intersection of the angle bisectors.
The barycentric coordinates of $I$ are $(a:b:c)$, with usual notations, bary-short.pdf, so we may write
$$
I =\frac1{a+b+c}(aA+bB+cC)\ .
$$
Now we need to find on the segment $AI$ the point $D$ with distance $k\sqrt 2$ to $A$. The distance $AI$ from $A$ to $I$ can be computed, and the answer is 
$$ 
D = A+\frac{k\sqrt2}{AI}(I-A)
=A+\frac{k\sqrt2}{AI}\cdot \frac1{a+b+c}(b(B-A)+c(C-A))
\ .
$$
Alternatively, let $A'$ be the point of intersection of $AI$ with the opposite side $BC$. The length of the angle bisector in $A$, $AA'$, can be computed. Then we may use $A'(0:b:c)$ instead of $I$ alternatively for the computation of $D$. We get
$$ 
D = A+\frac{k\sqrt2}{AA'}(A'-A)
= A+\frac{k\sqrt2}{AA'}\cdot \frac1{b+c}(b(B-A)+c(C-A))
\ .
$$
To get the final result, just plug in instead of $A,B,C$ their coordinate tuples, 
consider each tuple as a vector (in the vector space $\Bbb R^2$) and perform computations vectorially.
