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 $$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$
Best Answer
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.