Question

In the given figure, quadrilateral PQRS circumscribes the circle with centre O. Prove that the opposite sides of the quadrilateral PQRS subtend supplementary angles at the centre O.

Answer 1

Given: A quadrilateral PQRS circumscribing a circle with center O. Let the circle touch the sides PQ, QR, RS and SP at points A, B, C and D respectively.

To Prove:

1. ∠POQ + ∠ROS = 180°
2. ∠QOR + ∠SOP = 180°

Construction: Join the center O to the vertices P, Q, R, S and to the points of contact A, B, C, D,

Proof:

1. Triangle congruence:

Consider ΔOAP and ΔODP.

OA = OD   ...(Radii of the same circle)

PA = PD   ...(Tangents from an external point P are equal in length)

OP = OP   ...(Common side)

Therefore, ΔOAP ≅ ΔODP by the SSS congruence criterion.

By CPCT   ...(Corresponding parts of congruent triangles), ∠AOP = ∠DOP.

2. Naming the angles:

Let’s label the eight angles formed at the center O as shown in the diagram below:

Let ∠AOP = ∠1 and ∠DOP = ∠8.

Since ΔOAP ≅ ΔODP, ∠1 = ∠8.

Similarly, we can prove:

∠2 = ∠3   ...(From ΔOAQ ≅ ΔOBQ)

∠4 = ∠5   ...(From ΔOBR ≅ ΔOCR)

∠6 = ∠7   ...(From ΔOCS ≅ ΔODS)

3. Sum of angles at a point:

The sum of all angles around the point O is 360°:

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

4. Substitution:

Using the equalities from step 2, we substitute to group angles of opposite sides:

(∠1 + ∠1) + (2 + ∠2) + (∠5 + ∠5) + (∠6 + ∠6) = 360°

2(∠1 + ∠2 + ∠5 + ∠6) = 360°

∠1 + ∠2 + ∠5 + ∠6 = 180°

5. Conclusion:

Since ∠1 + ∠2 = ∠POQ and ∠5 + ∠6 = ∠ROS, we have:

∠POQ + ∠ROS = 180°

Similarly, it can be shown that:

∠QOR + ∠SOP = 180°