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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 $(D, E, F)$.

sketch

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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.

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  • $\begingroup$ Thank you for the answer, but I need the exact formula of D, E and F. I tried to find these points but I cannot solved. If I can find one of these points, I probably find the other two. $\endgroup$ – Sukru Alatas Jan 16 '18 at 20:04
  • $\begingroup$ You can write it all out in terms of your coordinates. This is easier to do by hand than setting it in Mathjax. So, for example, the vector I call $\underline{a}$ is $$\left(\begin{matrix}x_b-x_a\\y_b-y_a\end{matrix}\right)$$ etcetera $\endgroup$ – David Quinn Jan 17 '18 at 12:35

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