Given: 
ABCD is a parallelogram
AE bisects ∠BAD
BF bisects ∠ABC
CG bisects ∠BCD
DH bisects ∠ADC
To prove:
LKJI is a rectangle

 

I will just give an outline of how to prove it. You will have to
write it up as a two-column proof:

∠BAD + ∠ABC = 180° because adjacent angles of a parallelogram are supplementary

∠BAJ = ∠BAD because AE bisects ∠BAD       

∠ABJ = ∠ABC because DH bisects ∠ABC

∠BAJ + ∠ABJ = 90°  halves of supplemetary angles are complementary

ᐃABJ is a right triangle because its acute interior angles are complementary

Similar use ᐃCDL to prove ∠DLC = 90°

Similarly use ᐃADI to prove ∠AID = 90°

Then ∠JIL = 90° because ∠AID and ∠JIL are vertical angles

Then since 3 angles of quadrilateral LKJI are right angles, so
is the 4th one and so LKJI is a rectangle, since its interior 
angles are all right angles.

Edwin