Question

prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.....♡


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Answer 1

Answer:

step - by - step Explanation

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

o

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

o

=2(1+2+3+4)=360

o

=1+2+3+4=180

o

In ΔAOB,∠BOA=180−(1+2)

In ΔCOD,∠COD=180−(3+4)

∠BOA+∠COD=360−(1+2+3+4)

=360

o

–180

o

=180

o

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

Answer 2

Step-by-step explanation:

∆A0P Congruent to ∆A0S[As AP=AC,OP=OS,AO=AO Common]

Angle 1=Angle 8

2=3

4=5

6=7 ....... (i)

==>1,2,3,4,5,6,7,8=360°

...==>2Angle1+2Angle2+2Angle6+2Angle5=360°

..==>(Angle1+Angle2)+(Angle6+Angle 5)

==>Angle AOB+Angle COD=180°

Similarly,

Angle AOD+Angle BOC=180°.