Question

What is the value of the angle formed by intersection of bisectors of any two adjacent angles of a rhombus ? *
1 point
90 degree
25 degree
30 degree
105 degree​

Answer 1

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+∠D=180°.

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

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

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

Hence angle AOD=180°−((1/2)∠A+(1/2)∠D

=180°−90°=90°

Answer = 1