Obtuse angle bisector D of parallelogram ABCD intersects side AB at point K. Find the sides and angles of the parallelogram if KB is one and a half times greater than AK.
1) Let’s calculate the angles of the parallelogram.
The bisector divides the parallelogram into two figures: triangle AKD and trapezoid DKBC.
In a trapezoid, two angles adjacent to one side add up to 180 degrees. Let’s calculate ∠КDС = 180 – 134 = 46. ∠КDС = ∠КDA = ∠КDA = 46.
In the triangle AKD, we find the angle KAD:
∠КАD = 180 – 46 – 46 = 88;
∠ADC = 460 + 460 = 92;
Answer: ∠АDС = ∠ABС = 92; ∠BAD = ∠BCD = 88.
2) Calculate the sides of the parallelogram.
Triangle AKD is isosceles, since its two angles are equal, so AK = AD. The AB side is 2.5 parts, the AD side is 1 part.
Perimeter 5.6 cm = 1 piece + 1 piece + 2.5 pieces + 2.5 pieces = 7 pieces. 1 piece = 0.8 cm.
AD = BC = 0.8 cm;
DC = AB = 2 cm.
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