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Grade 11,

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ABCD is a cyclic parallelogram. :. L A + L C = 180° ...(i) [Sum of opposite angles of a cyclic quadrilateral is 180°.] Also, L A = L C ...(ii) [Opposite angles of a parallelogram] A 2L A = 180° L A= 90° As in a parallelogram one angle is 90°, hence it is a rectangle.

ABCD is a cyclic parallelogram.
:. L A + L C = 180° ...(i)
[Sum of opposite angles of a cyclic quadrilateral is 180°.]
Also, L A = L C ...(ii)
[Opposite angles of a parallelogram] A
2L A = 180° L A= 90°
As in a parallelogram one angle is 90°, hence it is a rectangle.

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
3 years ago
L ADE = L ABE ...(i) [Angles in the same segment of a circle] L ADF = L ACF ...(ii) [Reason same as above} Now L EDF = L ADE + L ADF B C L EDF = L ABE + L ACF D ...(iii ) [From (i), (ii)] As BE and CF are bisectors of L ABC and L ACB. 1 1 :. L ABE = 2 L ABC and L ACF = 2 L ACB ...(iv) From (iii) and (iv), we get 1 1 1 L EDF = 2 L ABC + 2 L ACB = 2 (L ABC + L ACB) In triangle ABC, L ABC + L ACB + L BAC = 180° .•.(v) 166 MATHEMATICS- I X L ABC + L ACB = 180° - L BAC. Substituting in (v), we get LEDF = (180° - LBAC) = 90° - 2 LBAC = 90°- 2 LA Similarly, we can show that LDEF = 90° - 2 LB; LDFE = 90° - 2 LC.

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