Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle
In the above figure, P, Q, R, S are points of contact
AS = AP (The tangents drawn from an external point to a circle are equal.)
∠SOA = ∠POA = ∠1 = ∠2 (Tangents drawn from a point outside of the circle, subtend equal angles at the centre)
∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8
Since complete angle is 360° at the centre,
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
2 (∠1 + ∠8 + ∠4 + ∠5) = 360° (or) 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°
∠1 + ∠8 + ∠4 + ∠5 = 180° (or) ∠2 + ∠3 + ∠6 + ∠7 = 180°
From above figure,
∠1 + ∠8 = ∠AOD, ∠4 + ∠5 = ∠BOC and ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD
Thus we have,
∠AOD + ∠BOC = 180° (or) ∠AOB + ∠COD = 180°
∠AOD and ∠BOC are angles subtended by opposite sides of quadrilateral circumscribing a circle and the sum of the two is 180°.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 13
It has been proved that the opposite sides of a quadrilateral ABCD circumscribing a circle with centre O subtend supplementary angles at the centre of the circle.
☛ Related Questions:
- In Figure 10.13, XY and X′ Y′ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X′ Y′ at B. Prove that ∠AOB = 90°.
- Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
- Prove that the parallelogram circumscribing a circle is a rhombus.
- A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.