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
