The bisector of angle A of triangle ABC divides side BC in half. Find side BC if AC = 3

The bisector of angle A of triangle ABC divides side BC in half. Find side BC if AC = 3 and the perimeter of triangle ABC is 10.

In this problem, we use the property of the median (divides the side of a triangle in half) and the theorem that if the bisector in a triangle is also the median, then such a triangle will be isosceles. Then we use the triangle perimeter formula.
1) According to the condition of the problem, the angle CAD = the angle of DAB and the side of CD = DВ, hence the triangle CAB is isosceles and for it AC = AB;
2) Since AC = 3, then AB = 3;
3) From the formula for the perimeter of the triangle P = AC + AB + BC, we express BC:
BC = P – AC – AB, i.e. BC = 10 – 3 – 3 = 4
Answer: BC = 4