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(B) Given the relations $c - a = r(b - a)$ and $w - u = r(v - u)$.Let $r = \lambda e^{i\alpha}$,where $\lambda = |r|$ and $\alpha = \arg(r)$.From the

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(B) Given the relations $c - a = r(b - a)$ and $w - u = r(v - u)$.Let $r = \lambda e^{i\alpha}$,where $\lambda = |r|$ and $\alpha = \arg(r)$.From the first relation,$c - a = \lambda e^{i\alpha}(b - a)$. Taking the modulus,$|c - a| = |r| |b - a|$,which implies $AC = |r| AB$.Taking the argument,$\arg(c - a) - \arg(b - a) = \arg(r) = \alpha$,which implies $\angle CAB = \alpha$.Similarly,from the second relation $w - u = r(v - u)$,we get $DF = |r| DE$ and $\angle FDE = \alpha$.Since $\frac{AC}{AB} = \frac{DF}{DE} = |r|$ and $\angle CAB = \angle FDE = \alpha$,by the $SAS$ similarity criterion,the triangles are similar.

If $a, b, c$ and $u, v, w$ are complex numbers representing the vertices of two triangles such that $c = (1 - r)a + rb$ and $w = (1 - r)u + rv$,where $r$ is a complex number,then the two triangles

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