Question

if (x-z) : (y-z) :: x² : y²then show that (x+y):(y+z) = (x/y + z) : (y/x + z) : (x≠y)​

Answer 1

Step-by-step explanation:

Given:

(x-z) : (y-z) :: x² : y²

=> (x-z)/(y-z) = x² /y²

=>(x-z)/x²  = (y-z)/y²

=>1/x - z/x² = 1/y - z/y²

=>1/x - 1/y = z/x² - z/y²

=>(x - y)/x*y = z( x² - y²)/ x²*y²

=>(x-y) = z(x-y)*(x +y)/x*y

=> 1 = z(x +y)/x*y

=>(x*y)/z = (x + y)

=> x/z + y/z = x + y

=> x/z - x = y - y/z