MHB Equlateral Triangle: Prove $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$

[lfdahl]

Given a triangle with angles $\alpha, \beta, \gamma$ opposite respectively to the sides $a,b,c$. Show in at least two different ways, that the triangle is equilateral if and only if $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$.

 

[kaliprasad]

Given a triangle with angles $\alpha, \beta, \gamma$ opposite respectively to the sides $a,b,c$. Show in at least two different ways, that the triangle is equilateral if and only if $ab \cos \gamma = ac \cos \beta = bc \cos \alpha$.

Method 1)

Spoiler

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
hence $\frac{1}{c}\cos \gamma =\frac{1}{b} \cos \beta = \frac{1}{a} \cos \alpha$
we have from sin rule $\frac{1}{c}\sin \gamma =\frac{1}{b} \sin \beta = \frac{1}{a} \sin \alpha$
square and add to get $\frac{1}{c^2} =\frac{1}{b^2} = \frac{1}{a^2} $ or $a=b=c$ hence equilateral

for the other part if $a=b=c$ then
$ \alpha = \beta = \gamma$
hence $ \cos \alpha = \cos \beta = \cos \gamma$ and by multiplying we get the result

method 2

using the values of cos we get

Spoiler

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
$<==> a^2+b^2-c^2 = a^2+c^2-b^2= b^2 + c^2 - a^2$
from $a^2+b^2-c^2 = a^2+c^2-b^2$
$<==> b^2-c^2<==>b=c$ simlilarly a =c $<==>$ it is equilateral

 

[lfdahl]

Method 1)

Spoiler

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
hence $\frac{1}{c}\cos \gamma =\frac{1}{b} \cos \beta = \frac{1}{a} \cos \alpha$
we have from sin rule $\frac{1}{c}\sin \gamma =\frac{1}{b} \sin \beta = \frac{1}{a} \sin \alpha$
square and add to get $\frac{1}{c^2} =\frac{1}{b^2} = \frac{1}{a^2} $ or $a=b=c$ hence equilateral

for the other part if $a=b=c$ then
$ \alpha = \beta = \gamma$
hence $ \cos \alpha = \cos \beta = \cos \gamma$ and by multiplying we get the result

method 2

using the values of cos we get

Spoiler

$ab \cos \gamma = ac \cos \beta = bc \cos \alpha$
$<==> a^2+b^2-c^2 = a^2+c^2-b^2= b^2 + c^2 - a^2$
from $a^2+b^2-c^2 = a^2+c^2-b^2$
$<==> b^2-c^2<==>b=c$ simlilarly a =c $<==>$ it is equilateral

Well done, kaliprasad! Thankyou for your participation!(Cool)