Question 536051
<pre>
Given: 
ABCD is a parallelogram
AE bisects &#8736;BAD
BF bisects &#8736;ABC
CG bisects &#8736;BCD
DH bisects &#8736;ADC
To prove:
LKJI is a rectangle

{{{drawing(400,2000/11,-1,21,-1,9,

line(0,0,15,0), line(15,0,39/2,sqrt(60.75)), line(39/2,sqrt(60.75),9/2,sqrt(60.75)), line(9/2,sqrt(60.75),0,0), green(line(27/2,sqrt(60.75),0,0),
line(6,0,39/2,sqrt(60.75)),line(9,0,9/2,sqrt(60.75)), line(21/2,sqrt(60.75),15,0)),

locate(0,0,A), locate(6,0,G), locate(9,0,H), locate(15,0,B),
locate(4.5,8.8,D),locate(10.5,8.8,F),locate(13.5,8.8,E),locate(19.5,8.8,C),
locate(6.7,5,I),locate(12.7,5,K),locate(8.1,2.5,L),locate(10.8,6.1,J)
 


   )}}} 

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

&#8736;BAD + &#8736;ABC = 180° because adjacent angles of a parallelogram are supplementary

&#8736;BAJ = {{{1/2}}}&#8736;BAD because AE bisects &#8736;BAD       

&#8736;ABJ = {{{1/2}}}&#8736;ABC because DH bisects &#8736;ABC

&#8736;BAJ + &#8736;ABJ = 90°  halves of supplemetary angles are complementary

&#5123;ABJ is a right triangle because its acute interior angles are complementary

Similar use &#5123;CDL to prove &#8736;DLC = 90°

Similarly use &#5123;ADI to prove &#8736;AID = 90°

Then &#8736;JIL = 90° because &#8736;AID and &#8736;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</pre>