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