In parallelogram ABCD, the bisector of angle A intersects side BC at point E.

In parallelogram ABCD, the bisector of angle A intersects side BC at point E. The BAE angle is 53 degrees. Find the smaller angle of the parallelogram ABCD

1. Let us denote the angle by the symbol ∠.

2. The bisector AE of the parallelogram ABCD separates the triangle ABE, which is isosceles, from it.

Therefore, ∠BAE = AEB = 53 °.

3. ∠BAE = ∠DAE = 53 °, since AE is a bisector.

4.∠А = ∠ВАЕ + ∠DАЕ = 53 ° + 53 ° = 106 °.

5.∠А = ∠С = 106 °.

4. We calculate the value of ∠D, taking into account that the total value of the angles adjacent to one of the sides of the parallelogram is 180 °:

∠D = 180 – ∠A = 180 ° – 53 ° = 127 °.

∠Д = ∠В = 127 °.

Answer: ∠А = ∠С = 106 ° – smaller angles of the parallelogram.