2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I did this can I delete it now if yes how to delete posts in se. $\endgroup$
    – user765842
    Commented Apr 13, 2020 at 10:58
  • $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Apr 13, 2020 at 11:01
  • 1
    $\begingroup$ I'm voting to close this question as off-topic because OP wants to delete it. $\endgroup$
    – Blue
    Commented Apr 13, 2020 at 11:02
  • $\begingroup$ Delete button is not there I am using Android $\endgroup$
    – user765842
    Commented Apr 13, 2020 at 11:03

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.