In triangle ABC: AB = BC, AD- median. The difference between the perimeters of the triangles ACD

In triangle ABC: AB = BC, AD- median. The difference between the perimeters of the triangles ACD and ADB is 2m, AB = 8. Find the AC.

Given: AB = BC = 8 m, BD = CD, PΔАСD – PΔАDВ = 2 m. It is required to determine AC.
As you know, the sum of the lengths of all sides is called the perimeter of a triangle. It is clear that the perimeter of the triangle АСD is equal to PΔАСD = АС + АD + СD. Similarly, the perimeter of triangle ADB is equal to P∆ADB = AD + BD + AB.
According to the setting condition PΔАСD – PΔАDВ = АС + АD + СD – (АD + BD + AB) = 2 m.Considering AB = 8 m and BD = CD, we have АС + АD + СD – АD – СD – AB = 2 m or AC – 8 m = 2 m, whence AC = 2 m + 8 m = 10 m.
Answer: AC = 10 cm.