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

9th Class Maths – NCERT Important Questions

Ans:

Step 1:

Let’s make a diagram for better understanding of the question:

Here, side AB subtends ∠ AOB at center and its opposite side CD subtends ∠ COD at center O.

Similarly, side BC subtends ∠ BOC at center and its opposite side AD subtends ∠ AOD at center O.

We need to prove that ∠ AOB + ∠ COD = 180⁰ and ∠ BOC + ∠ AOD = 180⁰

Step 2: 

Let’s compare Δ BPOand Δ BQO: 

OP = OQ                    (Radii of circle)

∠BPO = ∠ BQO         (Radius Ʇ tangent)

BO = BO                    (common side)

∴ Δ BPO Δ BQO

Now by Corresponding Parts of Congruent Triangles (CPCT) rule:

∠BOP = ∠ BOQ  ……………..(i)

Step 3: 

In the full diagram, it can be written as: O₁ = O₂

Let’s look at similar possibilty in the diagram, we get:

O₃ = O₄, O₅ = O₆ and O₇ = O₈

Step 4: 

We know that, sum of angles on a point is 360

∴ O₁ + O₂ + O₃ + O₄ + O₅ + O₆ + O₇ + O₈ = 360⁰

∴ (O₁ + O₂) + (O₃ + O₄) + (O₅ + O₆) + (O₇ + O₈) = 360⁰

∴ 2 O₁ + 2 O₄ + 2 O₅ + 2 O₈ = 360⁰               (by applying relations found in Step 3)

∴ O₁ + O₄ + O₅ + O₈ = 180⁰ 

∴ (O₁ + O₈ ) + (O₄ + O₅) =  180⁰ 

∴ ∠ AOB + ∠ COD =  180^{0   …………..…} Hence Proved !

Similarly, we can prove that ∠ BOC + ∠ AOD =  180⁰   

Therefore, theangles subtended by opposite sides are supplementary.

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