The bisector AD is drawn in triangle ABC, with AB = AD = CD. Find the smaller angle of triangle ABC.

Since, by condition, AB = AD = CD, the bisector AD splits the ABC triangle into two isosceles triangles: ABD and ACD.

Let the angle BAC = 2 * X0, then the angle BAD = CAD = X0.

In an isosceles triangle ACD, the angle ACD = CAD = X0.

In triangle ABD, angle BAD = DAB = (180 – X) / 2 = (90 – X / 2) 0.

The sum of the interior angles of the triangle ABC is 180, then:

180 = 2 * X + X + 90 – X / 2.

90 = 2.5 * H.

X = 36.

Angle ACD = 36, angle BAC = 2 * 36 = 72, angle ABC = 90 – 18 = 62.

Answer: The smallest angle of the triangle is 36.