If `x/a = y/b = z/c `, then show that ` (x^(2)-yz)/(a^(2)-bc) = (y^(2)-zx)/(b^(2) - ca) = (z^(2)-xy)/(c^(2)-ab)`

Explanation

` x/a = y/b = z/c = k " (let)" [k ne 0]`
` :. X = ak , y = bk " and " z = ck`.
Now, `(x^(2)-yz)/(a^(2)-bc) = ((ak)^(2)-(bk)(ck))/(a^(2)-bc)`
` = (a^(2)k^(2)-bck^(2))/(a^(2) - bc) = (k^(2)(a^(2)-bc))/((a^(2)-bc)) = k^(2) ` ............(1)
`(y^(2) - zx)/(b^(2) - ca) = ((bk)^(2)-ck.ak)/(b^(2) - ca) = (b^(2)k^(2)-k^(2)ca)/(b^(2)-ca) =(k^(2)(b^(2)-ca))/(b^(2)-ca) =k^(2) `...............(2)
`(z^(2)-xy)/(c^(2)-ab) = ((ck)^(2)-(ak)(bk))/(c^(2)-ab)= (c^(2)k^(2)-abk^(2))/(c^(2)-ab) = (k^(2)(c^(2)-ab))/((c^(2) - ab)) = k^(2)` .....................(3)
` :. ` We get from (1) , (2) and (3) , `(x^(2)-yz)/(a^(2)-bc) = (y^(2)-zx)/(b^(2)-ca) = (z^(2)-xy)/(c^(2)-ab)`.