Need an elegant proof for $(q-r)x+(r-p)y+(p-q)z=0$ with given premises

Question

The question comes from Higher Algebra by Hall and Knight:

If $$\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}$$ then shew that $(q-r)x+(r-p)y+(p-q)z=0$

I did solve this problem eventually, but I do not like the approach. Here is my take on this question:

We have $$\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}$$ Reciprocating each term, we have $$\frac{q+r-p}{x}=\frac{r+p-q}{y}=\frac{p+q-r}{z}$$ Taking the first two terms, we have $$\frac{q+r-p}{x}=\frac{r+p-q}{y}$$ $$\implies (r-p)y+qy-rx-px+qx=0 \tag{1}$$ Similarly, from the second and third terms, we have $$(p-q)z+rz-py-qy+ry=0\tag{2}$$ and from the first and third terms, we have $$(q-r)x+px-qz-rz+pz=0\tag{3}$$

On adding these three equations $(1)$, $(2)$ and $(3)$, we achieve after some tedious manipulations:

$$2((q-r)x+(r-p)y+(p-q)z)=0$$ $$\implies (q-r)x+(r-p)y+(p-q)z=0$$

Q.E.D.

I am not quite content with this proof. It seems... inelegant. Is there a better approach? If so, what is it?

Answer 1

If $$ t = \dfrac{x}{q+r-p} = \dfrac{y}{r+p-q} = \dfrac{z}{p+q-r} $$ we have $$\eqalign{(q−r)x+(r−p)y+(p−q)z &= t \left((q-r)(q+r-p) + (r-p)(r+p-q) + (p-q)(p+q-r)\right)\cr &= t(A - B)\cr} $$ where $$ A = (q-r)(q+r) + (r-p)(r+p)+(p-q)(p+q) = q^2 - r^2 + r^2 - p^2 + p^2 - q^2 = 0$$ and $$B = (q-r) p + (r-p) q + (p-q) r = qp - rp + rq - pq + pr - qr = 0 $$

Answer 2

By the well known property of proportions $$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_1+a_2}{b_1+b_2}$$ we have

$$\frac{x}{q+r-p}=\frac{y}{r+p-q}=\frac{z}{p+q-r}= K\ne 0$$ We can do $$\frac{x}{q+r-p}=\frac{x(q-r)}{q^2-r^2-pq+pr}$$ and similarly for the other two fractions, then the summ of denominators is nul so we get $$\frac{(q-r)x+(r-p)y+(p-q)z}{0}=K$$ Thus necessarily the numerator must be equal to $0$ because $K$ is not infinite.