Q. 115.0( 2 Votes )
The bisectors of two adjacent angles of a parallelogram intersect at
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)
∠EAB + ∠EBA + ∠AEB = 180° (sum of the angles of a triangle is 180°)
90° + ∠AEB = 180°
∠AEB = 90°
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