Question

Prove that the bisectors of interior angles of a parallelogram form a rectangle.

Answer 1

Given:
ABCD is a parallelogram
AE bisects ∠BAD
BF bisects ∠ABC
CG bisects ∠BAD
DH bisects ∠ADC

To prove: LKJI is a rectangle

Proof :
∠BAD + ∠ABC = 180° ...[ adjacent angles of a parallelogram are supplementary ] 
∠BAJ = `1/2` ∠BAD ...[AE bisects  BAD ]
∠ABJ = `1/2` ∠ABC ... [DH  bisect ABC ]
∠BAJ + ∠ABJ = 90° ...[ halves of supplementary angles are complementary ]

ΔABJ is a right triangle because its acute interior angles are complementary.
Similarly
∠DLC = 90°
∠AID = 90°

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

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