In parallelogram ABCD, the angle ABC is 150 °. The bisector of angle B intersects side AD at point K. Find the magnitude of the angle BKD in degrees.
The bisector BK divides the angle ABC into two equal angles:
<ABC = <KBC = 150 ° / 2 = 75 °.
The angles KBC and BKD are internal one-sided angles obtained by cutting parallel sides with a bisector:
<KBC + <BKD = 180 °;
<BKD = 180 ° – <KBC = 180 ° – 75 ° = 105 °.
Answer: 105 °.
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