In an isosceles trapezoid, the diagonal is 120 degrees with the lateral side.

In an isosceles trapezoid, the diagonal is 120 degrees with the lateral side. The lateral side is equal to the smaller base. Find the corners of the trapezoid.

ABCD – isosceles trapezoid, AB = BC = CD, angle ACD = 120 degrees.
1. Angles BCA and CAD are equal as criss-crossing angles formed due to the intersection of two parallel lines (BC and AD) secant (AC). Let’s denote:
BCA angle = CAD angle = x.
2. Consider the triangle ABC: according to the condition AB = BC, then the triangle ABC is isosceles, AB and BC are the lateral sides, and AC is the base, Angles CAB and BCA are the angles at the base of an isosceles triangle, therefore:
angle CAB = angle BCA = x.
3. By the theorem on the sum of the angles of a triangle in a triangle ABC:
angle CAB + angle ABC + angle BCA = 180 degrees;
x + angle ABC + x = 180;
angle ABC = 180 – 2x.
4. Since ABCD is an isosceles trapezoid, the angles ABC and BCD are equal as the angle at the base of the BC:
angle ABC = angle BCD = 180 – 2x.
angle BCD = angle BCA + 120 degrees = x + 120.
Then:
180 – 2x = x + 120;
-3x = -60;
x = 60/3;
x = 20.
Angle BCD (angle C) = x + 120 = 20 + 120 = 140 (degrees).
Angle C = Angle B = 140 degrees.
5. In a triangle ACD: angle CAD = x = 20 degrees, angle ACD = 120 degrees, then according to the theorem on the sum of the angles of a triangle:
CAD angle + ACD angle + CAD angle = 180 degrees;
20 + 120 + CAD angle = 180;
angle CAD = 180 – 140;
angle CAD = 40 degrees.
Angle CAD = Angle D = Angle A = 40 degrees (since these are the angles at the base of the isosceles trapezoid AD).
Answer: angle A = 40 degrees, angle B = 140 degrees, angle C = 140 degrees, angle D = 40 degrees.