In a parallelogram KITE angle KIT = 75 KA is perpendicular on IT. KA and EI intersect atB where A is a point on IT. If EB=2KI then measure of angle KEB is

KA ^ IT
Ð I = 75°  Ð AKI = 15° 
KE || IT
\  ÐAKE = 90°
Let D be the mid point of EB
Since in a right angled D, mid point of the hypotenuse is
equidistant from the vertices
\ DK = DB = DE Let ÐABI = q= ÐDKB = q
17
In DDKB, ÐBDK = 180° – 2q
In  DIKD
IK = KD
\  ÐKID = ÐKDI = 180–2q
In DIKD, 180° – 2q+ 180° – 2q+ q+ 15° = 180°
\ 180° + 15° = 4q– q
195° = 3q
65° = q
ÐABI = 65°