The three external bisectors meet the sides on a line

Idea. Compute the three ratios with the bisector theorem and check that the product is $+1$.

Existence. The external bisector at $A$ is parallel to $BC$ iff $AB=AC$; since the sides are pairwise different, the three points exist.

The ratios. By the external bisector theorem,
$$\frac{DB}{DC}=\frac{c}{b},\qquad\frac{EC}{EA}=\frac{a}{c},\qquad\frac{FA}{FB}=\frac{b}{a},$$
where $a=BC,b=CA,c=AB$. The product is $1$.

Signs. Each of $D,E,F$ is *outside* the corresponding segment, so each signed ratio is positive. Thus the signed product is $+1$.

Menelaus. The standard form says $D,E,F$ are collinear iff $\frac{\overline{BD}}{\overline{DC}}\cdot\frac{\overline{CE}}{\overline{EA}}\cdot\frac{\overline{AF}}{\overline{FB}}=-1$. Since $\overline{DB}=-\overline{BD}$, etc., our product equals $-$ of this standard product; "our product $=+1$" is equivalent, so $D,E,F$ are collinear. $\blacksquare$

Remark. For *internal* bisectors the feet are inside the sides and the lines concur (Ceva); for *external* ones the feet are collinear (Menelaus).

Lesson. Ceva = concurrency of cevians, Menelaus = collinearity of points on the side lines; both reduce to a product of ratios equal to $\pm1$.