In trapezoid ABCD (AD, BC – bases) AB = BC, AC = CD, BC + CD = AD. Find the corners of the trapezoid.

Let a trapezoid ABCD be given (AD, BC are bases), in which AB = BC, AC = CD, BC + CD = AD. On the AD side, construct a point K in such a way that CK | | AB, then ABCK will be a parallelogram. It turns out that AK = BC by the parallelogram property, and since AD ​​= BC + CD = AK + KD, then KD = CD. We denote the value ∠САD = x, then in ∆ ACD by the property of angles at the base of an isosceles triangle ∠СDА = ∠САD = x.
In the parallelogram ABCK ∠SAK = ∠ACB = x as internal criss-crossing at BC | | AK and secant AC. In ∆ ABC, according to the property of angles at the base of an isosceles triangle ∠BAC = ∠BCA = x, then ∠ ABC = 180 ° – (∠BAC + ∠BCA) = 180 ° – 2 ∙ x, since the sum of the angles in the triangle is 180 °. We get ∠ BAD = ∠ BAC + ∠САD = x + x = 2 ∙ x. In ∆ CDK, according to the property of angles at the base of an isosceles triangle СКDSC = КСDKS = (180 ° – x): 2 = 90 ° – x / 2. In trapezium ABCD: ∠BCD = ∠BCA + ∠ASK + ∠KCD = x + x + 90 ° – x / 2 = 1.5 ∙ x + 90 °. Knowing that the sum of the angles in the quadrilateral is 360 °, we make the equation:
x + 2 ∙ x + 180 ° – 2 ∙ x + 1.5 ∙ x + 90 ° = 360 °; x = 36 °, then the angles of the trapezoid
∠СDА = 36 °; ∠ BAD = 72 °; ∠ ABC = 108 °; ∠BCD = 144 °.
Answer: the angles of the trapezoid are 72 °; 108 °; 144 °; 36 °.