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 °.
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