# 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