Question

 If `(y + z)/a = (z + x )/b = (x + y)/c` then show that `x/[b + c - a ] = y/[c + a - b] = z/(a + b - c)`

Answer 1

`(y + z)/a = (z + x)/b = (x + y)/ c` 

By invertendo,

`a/(y + z)= b /(z + x )= c/(x + y)` 

`a/(y + z)= b /(z + x)= c/(x + y) = (b + c - a)/(z + x + x + y - y - z)`   ...( Theorem of equal ratios)

⇒ `a/(y + z)= b /(z + x)= c/(x + y) = (b + c - a)/[2x]`    ...(1)

Now,

`a/(y + z)= b /(z + x )= c/(x + y) = [c + a - b]/[x + y + y + z - z -x]`    ...( Theorem of equal ratios)

⇒ `a/(y + z)= b /(z + x)= c/(x + y) = (c + b - a)/[2y]`    ...(2)

Also,

`a/(y + z)= b /(z + x)= c/(x + y) = [a + b - c]/[y + z + z + x - x - y ]`   ...( Theorem of equal ratios)

⇒ `a/( y + z)= b /(z + x)= c/(x + y) = (a + b - c)/[2z]`    ...(3)

From (1), (2) and (3), we have

`[b + c - a]/(2x) = [c + a - b]/( 2y) = (a + b - c)/(2z)`

⇒ `[b + c - a]/x = [c + a - b]/y = (a + b - c)/z`

by invertendo,

⇒ `x/[b + c - a]=  y/[c + a - b]=  z/(a + b - c)`