Question

Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circlw

Answer 1

LetABCDbe a quadrilateralcircumscribing 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 = 1eq
Also ∠ABO = ∠CBO [Since, BA and BC are tangents]
Let ∠ABO = ∠CBO = 2eq
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 quad. ABCD = 360°
⇒2(1 + 2 + 3 + 4) = 360°
⇒1 + 2 + 3 + 4 = 180°
In ΔAOB, ∠BOA= 180 – (a + b)
In ΔCOD, ∠COD = 180 – (c + d)
Angle BOA + angle COD = 360 – (a + b + c + d)
= 360° – 180°
= 180°
Hence 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

Answer:

given: quadrilateral circumscribing

to prove: the opposite side of quadrilateral circumscribing a circle supplementary angles at the centre of the circle .

construction: joined AO ,OB .,OC,OD

proof: angle DAO= angle BAO( TANGENT OF DA AND AB ) equation 1

simillary, angle ABO= angle CBO equation 2

similarly, angleBCO= angle DCO equation 3

similarly, angle CDO= angle ADO equation 4

adding equation 1,2,3,and 4

1+2+3+4+1+2+3+4=360

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

1+2+3+4=180

IN triangle DOC

angle DOC+DCO+COD=180(angle sum property)

equation 5

similarly, angle AOB+ABO+OAB=180 equation 6

adding equation 5and6

DOC+DCO+COD+AOB+ABO+OAB=180+180

NOW,angle DOC+angle AOB=180

HENCE PROOVE.....