DM and CK are the bisectors of the angles DC of the trapezoid ABCD with the bases AD and BC, find the angle between these bisectors.
Let the value of the angle ADC = α, then, since DO is the bisector of the angle ADC, the angle ODC = ADC / 2 = α / 2.
The sum of the angles at the lateral side of the trapezoid is 180, then the angle ВСD = (180 – α), and since CO is the bisector of the angle ВСD, then the angle СОD = ВСD / 2 = (180 – α) / 2 = (90 – α / 2) …
Consider a triangle OCD.
The sum of the interior angles of the triangle is 180, then the angle СОD = (180 – (90 – α / 2) – α / 2) = (180 – 90 + α / 2 – α / 2) = 90.
Answer: The angle between the bisectors is 90.
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