In a convex quadrilateral ABCD AB = BC, AD = CD, ∠B = 100 °, ∠D = 104 °. Find the angle A

Draw the diagonal AC.
Consider the triangle ABC.
Since AB = BC, it means that the triangle ABC is isosceles.
By the theorem on the sum of the angles of a triangle:
180 ° = ∠B + ∠BAC + ∠BCA.
180 ° = 100 ° + ∠BAC + ∠BCA.
80 ° = ∠BAC + ∠BCA.
By the property of an isosceles triangle, ∠BAC = ∠BCA, then
∠BAC = ∠BCA = 80 ° / 2 = 40 °.
The ACD triangle is also isosceles.
By similar calculations we get: 180 ° = 104 ° + ∠DAC + ∠DCA.
∠DAC + ∠DCA = 76 ° / 2 = 38 °
∠A = ∠BAC + ∠CAD = 40 ° + 38 ° = 78 °
Answer: 78