As usual, we could suppose that $|a| = |b| = |c| = 1$. The equality in the hypothesis is equivalent with $(a + b + c)^2 = ab + bc + ca$, and thus $(a + b + c)^2 = abc\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$, which is the same with $(a + b + c)^2 = abc(a + b + c)$. It is clear now that $|a + b + c| \in \{0, 1\}$.

If $|a + b + c| = 0$, then the orthocenter of the triangle with the vertices whose complex coordinates are $a$, $b$, $c$, coincides with its circumcenter, and thus the triangle is equilateral.

If $|a + b + c| = 1$, then $(a + b + c) \overline{(a + b + c)} = 1$, and thus
$$(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) = 1,$$
that is $(a + b)(b + c)(c + a) = 0$.
Then, at least one of the above parenthesis is zero, and thus the triangle has two antipodal vertices, that is it is a right angled triangle.