# The bisectors of two adjacent angles of a parallelogram intersect atA. 30°B. 45°C. 60°D. 90°

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°

2EAB + 2EBA = 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°

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