euclidean-geometrygeometrytriangles
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.
Consider the bisectors of two of the angles. Note that for any point on each bisector, the perpendicular distance to the two sides of the angle it bisects is equal, so at the intersection, the perpendicular distance to all three sides of the triangle is equal. Therefore this point is also on the third bisector line.
This intersection point is the center of the incircle - the common distance to each side is the radius of that circle.
First, you prove that external bisectors $AD$, $BD$, and inner bisector $CD$ intersect in on point in the same manner as with 3 inner bisectors. => Points $C$, $I$ and $D$ lie on the same line.
Quadrilateral $IADB$ has two opposite angles $\angle IAD=\angle IBD=90^\circ$, so $IABD$ is cyclic with the center of circle $O$ in the center of $ID$. => Points $O$, $I$ and $D$ lie on the same line.
Then you combine both statements together.
Best Answer
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.