Q. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. [ Can anyone answer this for me ? ]​

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.

ʏᴇ ʟᴏ ʙʜᴀɪʏᴀ ᴘʀᴏᴠᴇ ʜō ɢʏᴀ _ᴠ_

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

i think that you should ask google for the answer to that problem because if we don't know it google does