Question

in the parallelogram ABCD , the bisectors of adjacent angles A and B intersect each other at E. prove that Angle AEB= 90degree

Answer 1

in triangle AEB

A/2 + B/2 + E = 180 (ASP). (1)

As
AE is bisector of A, so EAB = A/2
BE is bisector of B, so EBA = B/2

Now
A + B = 180
Adjacent angles of parallelogram

>> (A + B)/2 = 90

Put in (1)
90 + E = 180

>> E = 90°

Hence, proved.

Answer 2

As AE is the bisector of A and is the bisector of B

Therefore , in triangle AEB we have

Therefore, A/2+ B/2 +E=180.......1

Now ,

A + B = 180

(opp. side of the plgm.)

A/2. + B/2 = 90

Put in 1

E + 90=180

E=90

HENCE PROVE