In an isosceles triangle ABC, the side AC is the base, angle C = 40 °, angle ABC = 100 °, BD is the median. Find the angles of triangle ABD.

Since the triangle ABC is isosceles, the angles at the base of the AC are equal: angle A = angle C = 40 °.
BD is the median, and in an isosceles triangle, the median, lowered from the top to the bottom, is also the bisector and height.
Therefore, BD is the median, height, bisector of the ABC angle.
This means that the angle ABD = 100 ° / 2 = 50 °
The BAD angle is also 40 °
Angle АDВ = 180 ° – 50 ° – 40 ° = 90 ° (this angle can be found in another way: we learned that ВD is height, and height is a perpendicular dropped at a right angle, and a right angle is 90 °).
Answer: angle ABD = 50 °, angle BAD = 40 °, angle ADB = 90 °.