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

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.

• I did this can I delete it now if yes how to delete posts in se.
– user765842
Commented Apr 13, 2020 at 10:58
• There should be a "delete" item (along with the "share", "cite", "edit", etc). In the meantime, I'm going to vote to close so that no one posts an answer.
– Blue
Commented Apr 13, 2020 at 11:01
• I'm voting to close this question as off-topic because OP wants to delete it.
– Blue
Commented Apr 13, 2020 at 11:02
• Delete button is not there I am using Android
– user765842
Commented Apr 13, 2020 at 11:03

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.