How to prove that these triangles will be similar using the given relation?

Question

$A$,$B$, $C$, $D$ and $E$ are points on the complex plane representing complex numbers $z_{1}$, $z_{2}$, $z_{3}$, $z_{4}$ and $z_{5}$ respectively. If $(z_{3}-z_{2})z_{4}=(z_{1}-z_{2})z_{5}$, then prove that $\triangle$$ABC$ and $\triangle$$ODE$ are similar. I know that in order to show that these triangles are similar, I need to show that the ratios $AB/OD$, $BC/DE$ and $CA/EO$ are equal. So I found out the ratios; $AB/OD$ is $(z_{2}-z_{1})/z_{4}$, $BC/DE$ is $(z_{3}-z_{2})/(z_{5}-z_{4})$ and $CA/EO$ is $(z_{3}-z_{1})/z_{5}$. But I could not show that these are equal, also I am not sure about the ratios whether they are correct. Please help.

Answer 1

Note that the distance between $A$ and $B$ is $|z_1-z_2|$. (Take the modulus.) The given relation implies $$ \frac {AB}{BC} = \frac{|z_1-z_2|}{|z_2-z_3|} = \frac{|z_4|}{|z_5|} = \frac{OD}{OE}\ . $$ ($B$ "comes twice" in the left most fraction, as $O$ does it in the right most one.) So the similarity concerns the two triangles written in the right order of their vertices, correspondingly: $$ \begin{aligned} &\Delta ABC\ ,\\ &\Delta DOE\ . \end{aligned} $$ Now we rewrite thew above equality of proportions, building ratios of lenghts of corresponding sides, $$ \frac {AB}{DO} = \frac{|z_1-z_2|}{|z_4|} = \frac{|z_2-z_3|}{|z_5|} = \frac{BC}{OE}\ , $$ and here we may possibly want to also involve the third ratio, $\frac{AC}{DE}$, and show it is equal to the above common value. (An other possibility is to take the angle, which is extracted from the argument of $\frac{z_1-z_2}{z_3-z_2}=-\frac {z_4-0}{z_5-0}$, a relation equivalent to the given one.) $$ \frac {AC}{DE} = \frac{|z_1-z_3|}{|z_4-z_5|} \ . $$ And indeed, there is the relation $$ \frac {AC}{DE} = \frac{|z_1-z_3|}{|z_4-z_5|} = \left|\frac{z_1-z_3}{z_4-z_5}\right| \overset{(!)}= \left|\frac{z_1-z_2}{z_4}\right| = \frac {AB}{DO} \ . $$ In the equality marked with $(!)$ we have the equality of the quantities inside the absolute value, equivalent to the given relation.

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Note: Strictly speaking, the order of the vertices in a triangle is a part of the data of a triangle, two triangles are then similar when they are similar "for the given order of the vertices" building the corresponding proportions. So the problem if both "false" (seen from the one side) and a "trap" (seen from the other side, where we have to be generous and guess the order).