One of the bisectors of the triangle is divided by the intersection point of the bisectors in a ratio of 7: 6, counting from the top

One of the bisectors of the triangle is divided by the intersection point of the bisectors in a ratio of 7: 6, counting from the top. Find the perimeter of the triangle if the length of the side of the triangle to which this bisector is drawn is 48

Let AD be the bisector described in the condition.
BC is a side equal to 48.
Consider the triangle ADC.
For this triangle, CO is a bisector,
By the property of the bisector:
AO / OD = AC / CD = 7/6
6 * AC = 7 * CD
Consider the triangle ABD.
For this triangle, BO is a bisector,
By the property of the bisector:
AO / OD = AB / BD = 7/6
6 * AB = 7 * BD
Add up the resulting equalities:
6 * AC + 6 * AB = 7 * CD + 7 * BD
6 (AC + AB) = 7 (CD + BD), CD + BD = BC = 48
6 (AC + AB) = 7 * 48
AC + AB = 56
PABC = AC + AB + BC = 56 + 48 = 104
Answer: PABC = 104