Similarity of triangles in argand plane

**kalyanram** · December 14th 2010, 05:11 AM

Given $A(\alpha), B(\beta), C(\gamma)$ and $A'(\alpha'), B'(\beta'), C'(\gamma')$ are the vertices of $\triangle ABC$ and $\triangle A'B'C'$ respectively then show that the triangles are directly similar if
$\displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0$

Sol:
I tried out by saying that the $\triangle ABC$ and $\triangle A'B'C'$ are similar if the ratio of their sides are proportional and the $\angle A = \angle A' , \angle B = \angle B' and \angle C = \angle C'$ so we have
$\frac {|\alpha - \beta|}{|\alpha' - \beta'|} = \frac {|\beta - \gamma|}{|\beta' - \gamma'|} = \frac {|\gamma - \alpha|}{|\gamma' - \alpha'|}$ -- (Eq 1)
$arg(\frac {\alpha - \beta}{\beta - \gamma}) = arg(\frac {\alpha' - \beta'}{\beta' - \gamma'}) , arg(\frac {\beta - \gamma}{\gamma - \alpha}) = arg(\frac {\beta' - \gamma'}{\gamma' - \alpha'}) and arg(\frac {\gamma - \alpha}{\alpha - \beta}) = arg(\frac {\gamma' - \alpha'}{\alpha' - \beta'})$ -- (Eq 2)

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

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

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

Can you let me know the mistake with my arguement.

Thanks and Regards,
~Kalyan.

**Opalg** · December 14th 2010, 12:50 PM

Originally Posted by kalyanram

Given $A(\alpha), B(\beta), C(\gamma)$ and $A'(\alpha'), B'(\beta'), C'(\gamma')$ are the vertices of $\triangle ABC$ and $\triangle A'B'C'$ respectively then show that the triangles are directly similar if
$\displaystyle\sum \alpha{({\beta'}-{\gamma'})} = 0$

Sol:
I tried out by saying that the $\triangle ABC$ and $\triangle A'B'C'$ are similar if the ratio of their sides are proportional and the $\angle A = \angle A' , \angle B = \angle B' and \angle C = \angle C'$ so we have
$\frac {|\alpha - \beta|}{|\alpha' - \beta'|} = \frac {|\beta - \gamma|}{|\beta' - \gamma'|} = \frac {|\gamma - \alpha|}{|\gamma' - \alpha'|}$ -- (Eq 1)
$\arg(\frac {\alpha - \beta}{\beta - \gamma}) = \arg(\frac {\alpha' - \beta'}{\beta' - \gamma'}) ,\ \arg(\frac {\beta - \gamma}{\gamma - \alpha}) = \arg(\frac {\beta' - \gamma'}{\gamma' - \alpha'})$ 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

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

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

Can you let me know the mistake with my argument.

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

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

**kalyanram** · December 14th 2010, 09:18 PM

Hi Opalg,

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

~Kalyan.

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