We have parallelogram ABCD and isosceles triangle ADE, with equal angles ∠ADE = ∠AED.
∠ADE + ∠AED + ∠EAD = 180°
2 ∠ADE + 52° = 180°
∠ADE = (180° -52°) / 2 = 64°
∠ADE and ∠BAD are alternate interior angles of a transversal through parallel lines, so they're congruent. So ∠BAD is
Answer: 64°
