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Can you show us what you have tried?

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i am stuck in the initial stage itself not getting how will i proceed ....:confusedCan you show us what you have tried?

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- Feb 7, 2012

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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 $z(ax+by-cz)$.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)

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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

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i tried....its not working out.i am unable to solve .please helpthanks for advice... i will try that way and see

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- #7

- Feb 7, 2012

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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.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 $z(ax+by-cz)$.

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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 ....

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- #9

I removed the question you tagged on, as it was posted word for word (including the attempt) by another user, and the new topic can be found here:thanks a lot ... i got it ....

http://mathhelpboards.com/pre-algebra-algebra-2/continued-proportion-problem-7531.html