Question

show that the bisected of the angles of parallelogram enclose a rectangle

Answer 1

Answer.

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.

Answer 2

answer ✨
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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.

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