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

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

Angle B of triangle ABC is twice the angle A. The bisector of angle B divides the AC side into parts AD = 6 cm and CD = 3 cm. Find the sides of the triangle ABC.
AC = AD + CD = 6 cm + 3 cm = 9 cm
angle b = 2 * angle a
then the Triangle ABD is isosceles since the angle b and angle a are equal
angle A is X degrees
hence BD = AD = 6 cm.

consider the triangle BCD
angle CBD = 2 * angle BСD
hence the triangles BCD and ABC are similar in three angles, then the aspect ratio is also equal to
AC / BD = BC / DC = 9/6 = BC / 3.
Hence = BC = 6 cm
AB = AC = 9 cm