# Given a parallelogram ABCD, in which BF (F belongs to AD) is the bisector of angle ABC.

**Given a parallelogram ABCD, in which BF (F belongs to AD) is the bisector of angle ABC. Calculate the degree measures of the angles of the parallelogram ABCD if the angle AFB = 62 degrees.**

By the property of a parallelogram, sides BC and AD are parallel. Consequently, the angle AFB is equal to the angle CBF = 62 ° (internal criss-crossing angles with parallel BC and AD and secant BF).

BF is the bisector of the ABC angle, so the ABF angle is also 62 °. Hence, the angle ABC = 62 ° + 62 ° = 124 °.

In a parallelogram, two angles adjacent to one side will add up to 180 ° (by the property of a parallelogram), which means angle ABC + angle BAD = 180 °:

ABC + BAD = 180 °,

BAD = 180 ° – ABC = 180 ° – 124 ° = 56 °.

In a parallelogram, the opposite angles are equal, so the angles of the parallelogram are equal:

A = 56 °,

B = 124 °,

C = 56 °,

D = 124 °.