Question 13

Circumscribed Quadrilateral: A quadrilateral with a circle inside it (incircle) that touches all four sides. Each side is a tangent to the circle.

Supplementary Angles: Two angles whose sum is 180°. This proof shows that opposite sides subtend supplementary angles at the centre.

Tangents from an External Point are Equal: Two tangents from the same external point to a circle are always equal. Each vertex of ABCD acts as an external point.

Angles around a point = 360°: All angles formed around any single point always add up to 360°. Used in Step 4 to set up the equation.

We will use congruent triangles and the sum of angles around the center to prove the statement.

Step 1 — Draw and label

Let's draw a quadrilateral ABCD. This quadrilateral circumscribes a circle. Let O be the center of the circle. The sides AB, BC, CD, DA touch the circle. Let the touching points be P, Q, R, S respectively. Let's join O to A, B, C, D. Also, join O to P, Q, R, S.

Step 2 — Prove triangle congruence

Consider triangles ΔOAP and ΔOAS. OP and OS are radii of the same circle. So, OP = OS. AP and AS are tangents from point A. Tangents from an External Point are Equal: AP = AS since both are tangents from A.

Tangents from an external point are equal. So, AP = AS. AO is common to both triangles. So, AO = AO. SSS Congruence Rule: If all three sides of one triangle equal the corresponding three sides of another, the triangles are congruent.

By SSS congruence rule, ΔOAP ≅ ΔOAS.

Step 3 — Deduce equal angles

CPCT: Corresponding parts of congruent triangles are equal.

Since ΔOAP ≅ ΔOAS, their corresponding angles are equal. So, ∠POA = ∠SOA. Let's label these as ∠1 = ∠8. Similarly, we can prove other congruences. ΔPOB ≅ ΔQOB, so ∠POB = ∠QOB. Let's label these as ∠2 = ∠3. ΔQOC ≅ ΔROC, so ∠QOC = ∠ROC. Let's label these as ∠4 = ∠5. ΔROD ≅ ΔSOD, so ∠ROD = ∠SOD. Let's label these as ∠6 = ∠7.

Step 4 — Sum of angles at center

The sum of all angles around the center O is 360°. So, we can write: $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$

Step 5 — Substitute and simplify for one pair

We know that $\angle 1 = \angle 8$, $\angle 2 = \angle 3$, $\angle 4 = \angle 5$, and $\angle 6 = \angle 7$. Let's substitute these into the sum of angles. We replace $\angle 8$ with $\angle 1$. We replace $\angle 3$ with $\angle 2$. We replace $\angle 4$ with $\angle 5$. We replace $\angle 7$ with $\angle 6$. So, the sum becomes: $\angle 1 + \angle 2 + \angle 2 + \angle 5 + \angle 5 + \angle 6 + \angle 6 + \angle 1 = 360^\circ$ Combine the like terms: $2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^\circ$ Factor out 2: $2(\angle 1 + \angle 2 + \angle 5 + \angle 6) = 360^\circ$ Divide both sides by 2: $\angle 1 + \angle 2 + \angle 5 + \angle 6 = 180^\circ$ Now, let's group these angles. ∠1 + ∠2 is the angle ∠AOB. ∠5 + ∠6 is the angle ∠COD. So, we have: $\angle AOB + \angle COD = 180^\circ$ This proves that sides AB and CD subtend supplementary angles.

Step 6 — Conclude for the other pair

Similarly, we can prove this for the other pair of opposite sides. We would show that ∠BOC + ∠DOA = 180°.

Answer

$\boxed{\text{Hence, opposite sides subtend supplementary angles at the centre.}}$