The heights AD and BK of isosceles triangle ABC meet at point H, angle AHB = 128 degrees. Find the angles of triangle ABC.

Let AB be the base of an isosceles triangle.

Let’s draw the height CM of the triangle ABC. Since triangle ABC is isosceles, the height CM is the median of triangle ABC, then AM = BM = AB / 2.

Triangles АНМ and ВНМ are rectangular, in which АМ = ВМ, and the leg of MН is common, then the triangles are equal in two legs. Then the angle MAН = MВН = (180 – 128) / 2 = 26.

The angle AНK is adjacent to the angle AНВ, then the angle AНK = 180 – 128 = 52, then in the ACН triangle the angle CAН = (180 – 90 – 52) = 38.

Then the angle KAM = KAН + MAН = 26 + 38 = 64, and therefore the angle ABC is also equal to 64.

Angle ACB = (180 – 64 – 64) = 52.

Answer: The angles of triangle ABC are equal to 52, 64, 64.