Show that the angle bisector of a parallelogram is rectangle
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Given: ABCD is a parallelogram. AE bisects ∠BAD. BF bisects ∠ABC. CG bisects ∠BCD and DH bisects ∠ADC
To prove: LKJI is a rectangle
∠BAD + ∠ABC = 180° because adjacent angles of a parallelogram are supplementary

 [Since sum of adjacent angles of a parallelogram are supplementary]
ΔABJ is a right triangle since its acute interior angles are complementary
Similar in ΔCDL we get ∠DLC = 90° and in ΔADI we get ∠AID = 90°
Then ∠JIL = 90° as ∠AID and ∠JIL are vertical opposite angles
Since three angles of quadrilateral LKJI are right angles, hence 4th angle is also a right angle.
Thus LKJI is a rectangle.
To prove: LKJI is a rectangle
∠BAD + ∠ABC = 180° because adjacent angles of a parallelogram are supplementary

 [Since sum of adjacent angles of a parallelogram are supplementary]
ΔABJ is a right triangle since its acute interior angles are complementary
Similar in ΔCDL we get ∠DLC = 90° and in ΔADI we get ∠AID = 90°
Then ∠JIL = 90° as ∠AID and ∠JIL are vertical opposite angles
Since three angles of quadrilateral LKJI are right angles, hence 4th angle is also a right angle.
Thus LKJI is a rectangle.
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