A triangle with a perimeter of 30 is divided by its bisector into two triangles

A triangle with a perimeter of 30 is divided by its bisector into two triangles, the perimeters of which are 16 and 24. Find the bisector of this triangle.

Let a triangle ABC be given, its perimeter is 30, a bisector BN is drawn in it, it divides the AC side into segments AH and HC, and AC = AH + HC, and triangle ABC into two triangles ABН with a perimeter p1 = 16 and СВН with a perimeter p2 = 24.
Perimeter of triangle ABC:
p = AB + BC + AC.
Perimeter of triangle AВН:
p1 = AB + BH + AH.
SVN triangle perimeter:
p2 = CB + BH + CH.
Add the perimeters p1 and p2:
p1 + p2 = AB + BH + AH + CB + BH + CH = AB + 2 * BH + AH + CB + CH, we will replace AC = AH + HC:
p1 + p2 = AB + 2 * BН + CB + AC.
Now we subtract p from this expression:
p1 + p2-p = AB + 2 * ВН + CB + AC- (AB + BC + AC) = AB + 2 * ВН + CB + AC-AB-BC-AC = 2 * ВН.
Let us express ВН from this expression:
BH = (p1 + p2-p) / 2 = 16 + 24-30 / 2 = 5.
Answer: the bisector is 5.