Question

 If `[by + cz ]/[b^2 + c^2] = [cz + ax]/[c^2 + a^2] = [ax + by]/[a^2 + b^2]` then prove that `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]`

By theorem of equal ratios,

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

⇒ `[ 2ax + 2by + 2cz ]/[ 2a^2 + 2b^2 + 2c^2] `

⇒ `[ 2(ax + by + cz) ]/[ 2(a^2 + b^2 + c^2)]`

⇒ `[(ax + by + cz) ]/[(a^2 + b^2 + c^2)] `

Again applying theorem of equal ratios, we get

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

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

⇒ `x/a = y/b = z/c`