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 $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$
On
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?
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.