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Solution


Q.

In the given figure, AB, BC, CD and DA are tangents to the circle with centre O forming a quadrilateral ABCD. Show that ∠AOB+∠COD=180°

Ans.

Given A quad. ABCD circumscribes a circle with centre O.

To Prove: ∠AOB+∠COD=180°

and ∠AOD+∠BOC=180°

Join OP, OQ, OR and OS.

We know that the tangent drawn from an external point of a circle subtends equal angles at the centre.

⇒∠1=∠2,  ∠3=∠4,  ∠5=∠6,  ∠7=∠8,

and ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°  [∠S at a Point]

⇒2(∠2+∠3)+2(∠6+∠7)=360°

2(∠1+∠8)+2(∠4+∠5)=360°

⇒∠2+∠3+∠6+∠7=180°

∠1+∠8+∠4+∠5=180°

⇒∠AOB+∠COD=180°

∠AOD+∠BOC=180°