If $a:b:c=d:e:f$, how to show that $\frac{(a+b+c)^2}{(d+e+f)^2}=\frac{(ab+bc+ca)}{(de+ef+df)}$?

Question

If $a:b:c=d:e:f$, how to show that $\frac{(a+b+c)^2}{(d+e+f)^2}=\frac{(ab+bc+ca)}{(de+ef+df)}$?

Answer 1

Let $\dfrac ad=\dfrac be=\dfrac cf=k$(say)

$$\dfrac{ab+bc+ca}{de+ef+fd}=\dfrac{k^2(de+ef+df)}{de+ef+df}=k^2\text{ if } de+ef+df\ne0$$

$$k=\dfrac ad=\dfrac be=\dfrac cf=\dfrac{a+b+c}{d+e+f}\text{ if } a+b+c\ne0$$