Question

6. A circle touches all the four sides of a quadrilateral. Prove that the angle subtended
at the centre by the sides of a quadrilateral are complimentary to each other.
Please give right answer​

Answer 1

Answer:

A circle the centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P,Q,R and S respectively.

To prove: ∠AOB+∠COD=180

o

and, ∠AOD+∠BOC=180

o

Construction: Join OP,OQ,OR and OS

Proof:

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴∠1=∠2,∠3=∠4,∠5=∠6 and ∠7=∠8

Now, ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360

o

⇒2(∠2+∠3+∠6+∠7)=360

o

and

2(∠1+∠8+∠4+∠5)=360

o

(∠2+∠3+)+(∠6+∠7)=180

o

and (∠1+∠8)+(∠4+∠5)=180

o

[∵∠2+∠3=∠AOB, ∠6+∠7=∠COD, ∠1+∠8=∠AOD \ and \ ∠4+∠5=∠BOC]

⇒∠AOB+∠COD=180

o

⇒∠AOD+∠BOC=180

o

solution