Question

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

Answer 1

Since all the sides are equal in a rhombus . When 2 diagonals cut the rhombus into 2 equal parts i.e.,start from one vertex and joined to the opposite vertex and forms 2 equal parts and whenever the diagonal meet each other at a mid point they always form 90 degrees.

Answer 2

Let ABCD be a rhombus with AB and DC be 1 pair of  opposite parallel sides and AD and BC be the other pair of opposite parallel sides.

(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