In a convex quadrilateral ABCD: AB = BC, AD = CD, angle B = 3 ﾟ, angle D = 39 ﾟ. Find angle A.

1) And so, to begin with, we need to draw an AC diagonal, because of Δ ACD and Δ ABC, isosceles t to, by condition, their sides are equal because ⦟ D = 39 C then ⦟ CAD + ACD = 180 – 39 = 141 C , then ⦟ АСD = CAD = 141/2 = 70.5 C.
and now consider Δ ABC:
because ⦟ B is equal to 3 C, then BAC + BCA = 180 – 3 = 177 degrees, according to the theorem on the sum of the angles of a triangle
since Δ is isosceles, then its angles are equal, then ⦟ BAC = BCA = 177/2 = 88.5 C
then ⦟ A is equal to the sum of ⦟ BAC and CAD, i.e. 88.5 + 70.5 = 159 C

Answer: ⦟ A = 159 C
2) AC, then Δ ACD and ABC are isosceles t to by condition their sides are equal because ⦟ D = 39 degrees then ⦟ CAD + ACD = 180 – 39 = 141 C, then ⦟ ACD = CAD = 141: 2 = 70 , 5 C.
3) Consider Δ ABC:
since ⦟B is equal to 3 C, then BAC + BCA = 180 – 3 = 177 C, according to the theorem on the sum of angles Δ
since Δ is isosceles, then its angles at the base are equal, then ⦟ BAC = BCA = 177: 2 = 88.5 C
then ⦟ A is equal to the sum of ⦟ BAC and CAD i.e. 88.5 + 70.5 = 159 C
Answer ⦟ A = 159 C by the sum of angles theorem