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Prove ratio given equation

jewel

New member
Nov 3, 2013
5
if

x/(b+c-a)=y/(c+a-b)=z/(a+b-c)

prove that..

x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: hi i am stuck up with this numerical of fraction

Can you show us what you have tried?

Showing your work allows our helpers to see where you are stuck and what mistake(s) you may be making, so that they can offer suggestions to get you going again.
 

jewel

New member
Nov 3, 2013
5
Re: hi i am stuck up with this numerical of fraction

Can you show us what you have tried?

Showing your work allows our helpers to see where you are stuck and what mistake(s) you may be making, so that they can offer suggestions to get you going again.
i am stuck in the initial stage itself not getting how will i proceed ....:confused :confused:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
if

x/(b+c-a)=y/(c+a-b)=z/(a+b-c)

prove that..

x(by+cz-ax)=y(cz+ax-by)=z(ax+by-cz)
Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.
 

jewel

New member
Nov 3, 2013
5
Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.

thanks for advice... i will try that way and see
 

jewel

New member
Nov 3, 2013
5
thanks for advice... i will try that way and see
i tried....its not working out.i am unable to solve .please help (Worried)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Let \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) First show that $x+y+z = \lambda(a+b+c).$ Then see if you can find an expression for $ax$ in terms of $x,y,z$ and $\lambda$. Do the same for $by$ and $cx$. Then you will be able to see if $x(by+cz-ax)$ is the same as $y(cz+ax-by)$ and $z(ax+by-cz)$.
There are probably several ways to attack this problem. The method I used was to start with \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) Then $$x = \lambda(b+c-a),$$ $$y = \lambda(c+a-b),$$ $$z = \lambda(a+b-c).$$ Add those equations to get $x+y+z = \lambda(a+b+c)$. Now subtract the first of those three displayed equations, giving $y+z = 2\lambda a$. Next, multiply that by $x$: $$xy+xz = 2\lambda ax.$$ In a similar way, $$yx + yz = 2\lambda by,$$ $$zx+zy = 2\lambda cz.$$ At this stage, take a look at what you are trying to prove, and see if you can get there.
 

jewel

New member
Nov 3, 2013
5
There are probably several ways to attack this problem. The method I used was to start with \(\displaystyle \frac x{b+c-a} = \frac y{c+a-b} = \frac z{a+b-c} = \lambda.\) Then $$x = \lambda(b+c-a),$$ $$y = \lambda(c+a-b),$$ $$z = \lambda(a+b-c).$$ Add those equations to get $x+y+z = \lambda(a+b+c)$. Now subtract the first of those three displayed equations, giving $y+z = 2\lambda a$. Next, multiply that by $x$: $$xy+xz = 2\lambda ax.$$ In a similar way, $$yx + yz = 2\lambda by,$$ $$zx+zy = 2\lambda cz.$$ At this stage, take a look at what you are trying to prove, and see if you can get there.

thanks a lot ... i got it ....:)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775