Angle B of triangle ABC is twice the angle A. The bisector of angle B divides AC into parts AD

Angle B of triangle ABC is twice the angle A. The bisector of angle B divides AC into parts AD = 6 cm and CD = 3 cm find the sides of triangle ABC

Let the angle BAD = α, then ABC = 2 * α. Because BD is a bisector, then ABD = DBC = α.

In triangle ADB, the angle BAD = ABD = α. From this it follows that triangle ADB is isosceles, therefore BD = AD = 6 cm.

By the property of the bisector:

AD / DC = AB / BC.

Those. the bisector divides the opposite side into parts proportional to the adjacent sides.

BC / AB = DC / AD = 1/2.

Let BC = x, then AB = 2 * x.

Let’s use the bisector formula:

BD = √ (AB * BC – AD * DC) = √ (2 * x ^ 2 – 6 * 3) = 6;

36 = 2 * x ^ 2 – 18;

x = 3 * √3;

BC = 3 * √3 (cm);

AB = 6 * √3 (cm);

AC = AD + DC = 9 (cm).

Answer: BC = 3 * √3 cm, AB = 6 * √3 cm, AC = 9 cm.