In the triangle BAC, AD is the bisector. Sides AB and BCE are equal. Angle C = 64 degrees. Find the angle BDA.

Since, according to the condition, AB = BC, then the triangle ABC is isosceles with the base AC, and therefore the angle BAC = BCA = 64.

Since AD is the bisector of the angle BAC, it divides it in half, then the angle BAD = CAD = BAC / 2 = 64/2 = 32.

The desired angle BDA is the outer angle of the triangle ACD.

The outer angle of a triangle is equal to the sum of its two inner angles that are not adjacent to it.

Then the angle BDA = CAD + ACD = 32 + 64 = 96.

Answer: The value of the BDA angle is 96.