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Prove that the bisector of any two adjacent angles of a rhombus form a right angled triangle.

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by pkcgsk

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by pkcgsk

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(1).Now consider parallel sides AB and DC.

(2).The side AD is the intersection with these parallel sides.

(3) angle A and angle D are the internal angles on the same side of the parallel lines AB, DC formed with the intersection AD.

Therefore angle A + angle D = 180.

(4). Dividing both sides by 2, (1/2) angle A + (1/2) angle D = (1/2)x180 = 90.

(5). Let bisectors of angle A and angle D meet at O.

Then in triangle AOD , (1/2)angle A + (1/2) angle + angle AOD = 180

Hence angle AOD = 180 - {((1/2) angle A + (1/2) angle D}

= 180 - 90 = 90