Originally Posted by

**kalyanram** Given $\displaystyle A(\alpha), B(\beta), C(\gamma)$ and $\displaystyle A'(\alpha'), B'(\beta'), C'(\gamma')$ are the vertices of $\displaystyle \triangle ABC$ and $\displaystyle \triangle A'B'C'$ respectively then show that the triangles are directly similar if

$\displaystyle \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0 $

Sol:

I tried out by saying that the $\displaystyle \triangle ABC$ and $\displaystyle \triangle A'B'C'$ are similar if the ratio of their sides are proportional and the $\displaystyle \angle A = \angle A' , \angle B = \angle B' and \angle C = \angle C'$ so we have

$\displaystyle \frac {|\alpha - \beta|}{|\alpha' - \beta'|} = \frac {|\beta - \gamma|}{|\beta' - \gamma'|} = \frac {|\gamma - \alpha|}{|\gamma' - \alpha'|}$ -- (Eq 1)

$\displaystyle \arg(\frac {\alpha - \beta}{\beta - \gamma}) = \arg(\frac {\alpha' - \beta'}{\beta' - \gamma'}) ,\ \arg(\frac {\beta - \gamma}{\gamma - \alpha}) = \arg(\frac {\beta' - \gamma'}{\gamma' - \alpha'})$ and $\displaystyle \arg(\frac {\gamma - \alpha}{\alpha - \beta}) = \arg(\frac {\gamma' - \alpha'}{\alpha' - \beta'})$ -- (Eq 2)

From Eq 1 and Eq 2 we can conclude that the complex numbers

$\displaystyle \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'} ,\ \frac {\beta - \gamma}{\gamma - \alpha} = \frac {\beta' - \gamma'}{\gamma' - \alpha'}$ and $\displaystyle \frac {\gamma - \alpha}{\alpha - \beta} = \frac {\gamma' - \alpha'}{\alpha' - \beta'}$ rearranging and solving this I cannot conclude that

$\displaystyle \displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0 $

Can you let me know the mistake with my argument.