# "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."

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.
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bargeolonakc
Let the triangle be $ABC$ and external angle bisector of $\mathrm{\angle }ABC$ cut $AC$ in $X$, of $\mathrm{\angle }ACB$ cut $AB$ in $Y$, of $\mathrm{\angle }BAC$ cut $BC$ in $Z$.
By angle bisector theorem,
$\frac{AX}{XC}=-\frac{AB}{BC}...\left(1\right)$
$\frac{CZ}{ZB}=-\frac{CA}{AB}...\left(2\right)$
$\frac{BY}{YA}=-\frac{BC}{CA}...\left(3\right)$
(1) ×(2) ×(3) gives,
$\frac{AX.CZ.BY}{XC.ZB.YA}=-1$
Therefore by converse of Menelaus Theorem X, Y, Z are collinear.