Question

In the figure, bisectors of $\angle$B and $\angle$C of $\Delta$ABC intersect each other in point X. Line AX intersects side BC in point Y. AB=$5$, AC=$4$, BC=$6$, then find $\frac{AX}{XY}$

Answer 1

Since, the bisectors of a triangle are concurrent, then $AY$ is the bisector of $\angle A$.

Then from angle-bisector property we get, [Using the bisector of $\angle C$ ]

$\frac{AX}{XY}=\frac{AC}{YC}$........(1).

Again, since $AY$ is the bisector of $\angle A$ then

$\frac{BY}{YC}=\frac{AB}{AC}$

or, $\frac{BY}{YC}=\frac{5}{4}$

or, $\frac{BY+YC}{YC}=\frac{5+4}{4}$

or, $\frac{BC}{YC}=\frac{9}{4}$

or, $YC=\frac{8}{3}$.

From (1) we get,

$\frac{AX}{XY}=\frac{3}{2}$.