Prove that external bisectors of the angles of a triangle meet the opposite sides in three collinear points.

Question

Prove that the external bisectors of the angles of a triangle meet the opposite sides in three collinear points.

I need to prove this using only Menelaus Theorem, Stewart's Theorem, Ceva's Theorem.

What I did:I tried by making a simple case diagram that is a diagram with obtuse angle in the given triangle. Then using Menelaus on angle bisectors with respect to the triangles and using angle bisector theorem for ratios of values.

Answer 1

Let the triangle be $ABC$ and external angle bisector of $\angle ABC$ cut $AC$ in $X$, of $\angle ACB$ cut $AB$ in $Y$, of $\angle BAC$ cut $BC$ in $Z$.

By angle bisector theorem, $$\frac{AX}{XC}=-\frac{AB}{BC}... (1)$$ $$\frac{CZ}{ZB}=-\frac{CA}{AB}... (2)$$ $$\frac{BY}{YA}=-\frac{BC}{CA}... (3)$$

(1) ×(2) ×(3) gives, $$\frac{AX.CZ.BY}{XC.ZB.YA}=-1$$ Therefore by converse of Menelaus Theorem X, Y, Z are collinear.