Question

Yah kar k dhikho......​

Answer 1

ANSWER

⇝ Let ABCD be a quadrilateral circumscribing a circle with centre O.

⇝ Now join AO, BO, CO, DO.

⇝ From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]

⇝ Let ∠DAO =∠BAO = 1

⇝ Also ∠ABO =∠CBO [Since, BA and BC are tangents]

⇝ Let ∠ABO =∠CBO = 2

⇝ Similarly we take the same way for vertices C and D

⇝ Sum of the angles at the centre is 360°

⇝ Recall that sum of the angles in quadrilateral, ABCD = 360°

= 2(1+2+3+4) = 360°

= 1 + 2 + 3 + 4 = 180°

⇝ In ΔAOB, ∠BOA = 180 − (1+2)

⇝ In ΔCOD, ∠COD = 180 − (3+4)

⇝ ∠BOA + ∠COD= 360 − (1+2+3+4)

= 360° – 180•

= 180°

⇝ Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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