Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angles.
Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angles.
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In parallelogram adjoining figure,
Given : ABCD is parallelogram with its bisector AE and BE meeting at E.
To Prove : < AEB = 90°
Proof : As ABCD is parallelogram so AD || BC where AB is transversal.
< A + < B = 180° [ Sum of interior consecutive angle ] ----(1)
On dividing equation 1 by 2 we get
< A / 2 + < B / 2 = 180° / 2
Since, AE and BE are angle biscetor
< A / 2 = < BAE and < B / 2 = < ABE
< BAE + < ABE = 90° -----(2)
In triangle AEB,
< BAE + < ABE + < AEB = 180° [ Angle sum property ]
90° + < AEB = 180° [ From (2) ]
< AEB = 180° - 90°
< AEB = 90°