Question

The bisectors of two adjacent angles of a parallelogram intersect at 

A. 30° 

B. 45° 

C. 60° 

D. 90°

Answer 1

Let ABCD is a parallelogram. 

The angle bisectors AE and BE of adjacent angles A and B meet at E. 

AD || BC (Opposite sides of ||gm) 

∠DAB + ∠CBA = 180° 

2∠EAB + 2∠EBA = 180° (sum of the interior angles, formed on the same side of the transversal, is 180°) 

AE and BE are the bisectors of ∠DAB and ∠CBA respectively. 

∠EAB + ∠EBA = 90° ... (1) 

In ∆EAB, 

∠EAB + ∠EBA + ∠AEB = 180° (sum of the angles of a triangle is 180°) 

90° + ∠AEB = 180° 

From (1) 

∠AEB = 90°