A triangle has corners at points A, B, and C. Side AB has a length of #9 #. The distance between the intersection of point A's angle bisector with side BC and point B is #4 #. If side AC has a length of #15 #, what is the length of side BC?

The bisector of an angle of a triangle divides the opposite side of the triangle into two segments whose ratio is the same as the ratio of the triangle's sides of the bisected angle.

In the diagram below,if #P# is the intersection of the the bisector of #/_CAB# with the edge #BC# then
#color(white)("XXX")abs(AC)/abs(AB)=abs(PC)/(abs(BP)# (see below if needed)

Given: #abs(AC)=15#, #abs(AB)=9#, and #abs(BC)=4#
we have
#color(white)("XXX")abs(PC)=15/9 xx 4/color(white)(1)=20/3=6 2/3#

#abs(BC)=abs(BP)+abs(PC)=4+6 2/3=10 2/3#

#color(blue)("=======================================================")#

In case you have not been given the ratio rule used above,
here is a quick proof of why it is true.

Draw a line parallel to #AB# through #C# and extend the bisector #AP# to intersect this line at #Q#

SInce #AB || CQ#
#color(white)("XXX")/_CQP=/_PAB#
#color(white)("XXX")/_QCP=/_ABP#
#rArr triangleQCP and traingleABP# are similar
#color(white)("XXX")rarr abs(CP)/abs(BP)=abs(CQ)/(abs(AB)#

Furthermore since #/_CQP=/_PAB# and #/_PAB=/_PAC#
#triangleACQ# is isosceles with #abs(AC)=abs(CQ)#

Therefore
#color(white)("XXX")abs(CP)/abs(BP)=abs(AC)/(abs(AB)#