How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic Class 9

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic Class 9

Question 1.

How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°? Class 9

Solution:

∆OAB, ∆OAD, ∆OCD and ∆OCB are isosceles triangles.

∠OBA = ∠OAB = p

∠OCB = ∠OBC = q

∠OAD = ∠ODA = v

∠ODC = ∠OCD = u

... [Angles opposite to equal sides]

∠A = ∠OAD + ∠OAB = v + p

∠B = ∠OBA + ∠OBC = p + q

∠C = ∠OCD + ∠OCB = u + v

∠D = ∠CDO + ∠ODA = u + v

Since sum of interior angles of a quadrilateral = 360°

∴ ∠A + ∠B + ∠C + ∠D = 360°

∴ (v + p) + (p + q) + (u + q) + (u + v) = 360° ... [From (i)]

∴ 2 (p + q + u + v) = 360°

∴ (p + q) + (w + v) = 180°

∴ ∠B + ∠D = 180° ...[From (i)]

Similarly, we can show that ∠A + ∠C = 180°

∴ Sum of opposite angles of a cyclic quadrilateral is 180°