$(x+y) prop z,(y+z) prop x$ then prove that $(z+x) prop y$ ?thanks

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Answer 1 · 2017-06-09T05:15:58

Given

$x+ypropz$

$=>x+y=mz.......[1]$ , where m = proportionality constant

$=>(x+y)/z=m$

$=>(x+y+z)/z=m+1 ....[2]$

Again

$y+zpropx$

$=>y+z=nx........[3]$ , where n = proportionality constant

$=>(y+z)/x=n$

$=>(x+y+z)/x=n+1 ......[4]$

Dividing [2] by [4]

$x/z=(m+1)/(n+1)=k(say)$

$=>x=kz......[5]$

By [1] and [5] we get

$kz+y=mz$

$=>y=(m-k)z$

$=>y/z=(m-k)......[6]$

Dividing [2] by [6] we get

$(x+y+z)/y=(m+1)/(m-k)=c " another constant"$

$=>(x+y+z)/y-1=c -1$

$=>(x+z)/y=c -1="constant"$

Hence

$z+xpropy$

Proved

(x+y) prop z,(y+z) prop x(x+y) prop z,(y+z) prop x then prove that (z+x) prop y (z+x) prop y ?thanks