Question

If `(by + cz)/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)`, prove that each of these ratio is equal to `x/a = y/b = z/c`

Answer 1

`(by + cz)/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)`

= `(2(ax + by + cz))/(2(a^2 + b^2 + c^2)`

= `(ax + by + cz)/(a^2 + b^2 + c^2)`  ...(Adding)

Now `(by + cz)/(b^2+ c^2) = (ax + by + cz)/(a^2 + b^2 + c^2)`

⇒ `(by + cz)/(ax + by + cz) = (b^2 + c^2)/(a^2 + b^2 + c^2)`  ...(By alternendo)

⇒ `(by+ cz - ax- by- cz)/(ax + by + cz)`

= `(b^2 + c^2 - a^2 - b^2 - c^2)/(a^2 + b^2 + c^2)`

⇒ `(-ax)/(ax + by + cz) = (-a)/(a^2 + b^2 + c^2)`

⇒ `x/(ax + by + cz) = a/(a^2 + b^2 + c^2)`

⇒ `x/a = (ax + by + cz)/(a^2 + b^2 + c^2)`   ...(i)
Similarly we can prove that

`y/b = "ax + by + cz"/(a^2 + b^2 + c^2)`    ...(ii)

and `z/c = "ax + by + cz"/(a^2 + b^2 + c^2)`  ...(iii)
from (i), (ii) and (iii)

Hence `x/a = y/b = z/c`.