Question

I can't solve this problem ​

Answer 1

Step-by-step explanation:

Let ABCD be a quadrilateral circumscribing a circle with center 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=2

Similarly we take the same way for vertices C and D

Sum of the angles at the center is 360°

Recall that sum of the angles in quadrilateral, ABCD =360°=2(1+2+3+4)=360°

=1+2+3+4=180°

In ∆AOB,BOC=180-(1+2)

In ∆COD, 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 center of circle.

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