CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


Solution

Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join $$AO$$, $$BO$$, $$CO$$, $$DO$$.

From the figure, $$\angle DAO = \angle BAO$$ [Since, AB and AD are tangents]

Let $$\angle DAO = \angle BAO = 1$$

Also $$\angle ABO = \angle CBO$$ [Since, BA and BC are tangents]

Let $$\angle ABO = \angle CBO = 2$$

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 $$\Delta AOB, \angle BOA= 180 - (1 +2)$$

In $$\Delta COD, \angle COD = 180 - (3 +4)$$

$$\angle BOA + \angle COD = 360 - (1 + 2 + 3+ 4)$$

$$= 360^o – 180^o$$

$$= 180^o$$

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.


553959_494207_ans_79e310cf3ae04dd68b6a3cbd8106eca6.png

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image