In rhombus ABCD, angle B is obtuse. On the AD side, point K is taken, such that BK is perpendicular to AD. straight lines BK and AC intersect at point O, AC = 2BK. find the angle AOB.
By condition, the diagonal AC = 2 * BK. Since the diagonals of the rhombus are halved at the intersection point, AM = AC / 2 = BK.
In the ABD triangle, the heights AM and BK are equal, then the ABD triangle is equilateral, and all its angles are 60.
In an equilateral triangle, the heights are also the bisectors of the angles, then the angle ABO = BAM = 30.
In the triangle AOB, the angle AOB = 180 – OAB – OBA = 180 – 30 – 30 = 120.
Answer: The angle OAB is 120.
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