Let $A B C$ be a triangle with $A B=13, B C=15$ and $A C=9$. Let $r$ be the line parallel to $B C$ drawn through $A$. The bisector of angle $A \widehat{B C}$ intersects the line $r$ at $E$ and the bisector of angle $A \widehat{C} B$ intersects $r$ at $F$. Calculate the measure of the segment $EF$.

##
Since the line EF is parallel to the side $\mathrm{BC}$, the alternate interior angles formed by the transversal CF are equal, that is, FĈB $=C \hat{F} A$. On the other hand, since CF is the angle bisector, we have FCEB $=F \hat{C A}$, and thus, $F \hat{C A}=C \hat{F} A$, which means triangle CAF is isosceles with base CF. Therefore, $A F=A C=9$.

Similarly, we conclude that triangle $B A E$ is isosceles with base $B E$ and $A E=A B=13$. Thus, $E F=E A+A F=22$.