Prove ratio given equation

jewel

New member
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

Staff member
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
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 Opalg

MHB Oldtimer
Staff member
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
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
thanks for advice... i will try that way and see
i tried....its not working out.i am unable to solve .please help Opalg

MHB Oldtimer
Staff member
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
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 .... 