# Thread: Similarity of triangles in argand plane

1. ## Similarity of triangles in argand plane

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

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

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

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

Can you let me know the mistake with my arguement.

Thanks and Regards,
~Kalyan.

2. 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.
What's the problem, you're practically there! If $\displaystyle \frac {\alpha - \beta}{\beta - \gamma} = \frac {\alpha' - \beta'}{\beta' - \gamma'}$ then (multiplying out the fractions) $\displaystyle (\alpha - \beta)(\beta' - \gamma') = (\beta - \gamma)(\alpha' - \beta')$. Multiply out the brackets, cancel the term $\displaystyle -\beta\beta'$ on both sides, and rearrange to get $\displaystyle \alpha(\beta' - \gamma') +\beta(\gamma'-\alpha') + \gamma(\alpha'-\beta')$, or in other words $\displaystyle \sum \alpha{({\beta'}-{\gamma'})} = 0$.

In fact, the question asked for the converse result: If $\displaystyle \sum \alpha{({\beta'}-{\gamma'})} = 0$ then the triangles are similar. But your argument is reversible, so you have proved that result too.

3. Hi Opalg,

Thanks for the comment. Yeah sometimes I do miss the obvious.

~Kalyan.