In triangle ABC, side AC = 15 cm. The tangent point of the inscribed circle divides side AB in proportion

In triangle ABC, side AC = 15 cm. The tangent point of the inscribed circle divides side AB in proportion to the numbers 2 and 1, starting from the vertex A. Find the sides of the triangle if its perimeter = 42 cm.

Let the points K, H, M be points of tangency of the circle. We denote the length of the segment BH = X cm, then AH = 2 * X cm.

By the property of a tangent drawn from one point, ВK = BH = X cm, AM = AH = 2 * X cm.

Then CM = CK = (15 – 2 * X) cm.

The perimeter of the triangle ABC is equal to: Ravs = ВK + ВН + AН + AM + CM + СK = X + X + 2 * X + 2 * X + 15 – 2 * X + 15 – 2 * X = 42.

2 * X = 42 – 30 = 12.

X = 12/2 = 6 cm.

AB = 3 * 6 = 18 cm.

BC = 42 – 15 – 18 = 9 cm.

Answer: The lengths of the sides of the triangle are 9 cm, 15 cm, 18 cm.