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Question
Prove that the bisectors of interior angles of a parallelogram form a rectangle.
Solution
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 4th one and so is LKJI a rectangle, since its interior angles are all right angles
Hence proved.
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