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If $z_1,z_2,z_3$ and $z_1',z_2',z_3'$ are the vertices of similar triangles, then $$\begin{vmatrix}1&1&1\\z_1&z_2&z_3\\z_1'&z_2'&z_3'\end{vmatrix}=0$$

Where does this condition comes from ?

I just know that the area of the triangle is $$\Delta=\begin{vmatrix}1&1&1\\z_1&z_2&z_3\\\bar{z}_1&\bar{z}_2&\bar{z}_3\end{vmatrix}$$

and similar triangles satisfy $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$

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  • $\begingroup$ If the two triangles are similar then one set of those vectors can be represented as a scaled translation of the other set of vectors (write them in such a way). You can then find linear dependence of the row vectors. $\endgroup$ – user328442 Feb 17 at 5:08
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You obtain $z_3-z_1$ from $z_2-z_1$ by means of the same rotation+homothety as you obtain $z_3'-z_1'$ from $z_2'-z_1'$ (by similarity). This implies that they are related by the same complex number $$ \frac{z_3-z_1}{z_2-z_1}=\frac{z_3'-z_1'}{z_2'-z_1'} $$ If you expand this you should arrive at your determinant condition, since the above condition contains all the information (of course similar relations hold if you take cyclic permutations of the vertices) I am assuming here that the two triangles have the same orientation.

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Are you sure your determinant for similar triangle condition correct because consider a triangle of vertices $z_1,z_2,z_3$ and triangle formed by there conjugate , reflection is an isometry hence they are similar but , the determinant being the area of either of the triangle do not vanishes if $z_1,z_2,z_3$ are linearly independent , So?

Where have I mistaken in my above argument?

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  • $\begingroup$ I assumed that the triangles have the same orientation I mentioned that in my answer). If they have opposite orientation, the given condition is not right. In that case, you would say that the rotation+homothety from $z_1z_2$ to $z_1z_3$ is the conjugate of the one from $z_1'z_2'$ to $z_1'z_3'$. Your condition will be similar, but you need to take conjugate on one of the sides. You would arrive at the same determinant with either$z_1$, $z_2$ and $z_3$ or $z_1'$, $z_2'$ and $z_3'$ replaced by their conjugates. $\endgroup$ – GReyes Mar 6 at 2:47
  • $\begingroup$ @GReyes thank you very much for clearing the doubt. $\endgroup$ – Bijayan Ray Mar 6 at 12:08

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