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# if a(y+z)=x , b(z+x)=y , c(x+y)=z prove that x^2/a(1-bc)=y^2/b(1-ca)=z^2/c(1-ab)

Arun
25763 Points
3 years ago
A(y+z)=b(z+x)=c(x+y)
Assume n= a(y+z)=b(z+x)=c(x+y)
n=a(y+z) ,n= b(z+x) , n= c(x+y)
y+z = (n/a).....................(1)
z+x = (n/b).....................(2)
x+y = (n/c).....................(3)
subtracting (2) - (1) , (3) - (2) , (1) - (3) we get
x - y = n( a - b )/ab ..................(4)
y - z = n ( b - c )/bc...................(5)
z - x = n( c - a )/ac...................(6)

x-y/c(a-b) = [ n ( a - b )/ab ] / c( a - b )
= n/abc
y-z/a(b-c) = [ n ( b - c )/bc ] / a( b - c )
= n/abc
z-x/b(c-a) = [ n ( c - a )/ac ] / b( c - a )
= n/abc
y-z/a(b-c)=z-x/b(c-a)=x-y/c(a-b)

also
x^2/a(1-bc)=y^2/b(1-ca)=z^2/c(1-ab)