Question

Prove that in a parallelogram, the bisectors of any two consecutive angles intersect at right angles.

Answer 1

$\textbf{ \huge{ Hello Mate! }}$

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°

$\textbf{ \large{ Q.E.D} }$

$\textbf{ Have Great future ahead! }$