Circles | Exercise 10.2

Question 13

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

Solution

Understand the Question

A quadrilateral $ABCD$ circumscribes a circle with centre $O$ , touching the four sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively.
Tangents drawn from an external point to a circle are equal in length, which allows us to prove pairs of adjacent triangles congruent by the SSS congruence criterion.
By CPCT, the angles subtended at the centre by the tangents from each vertex are equal.
Since the sum of all angles around the centre is $360^\circ$ , we substitute these equal angles to show that opposite sides subtend supplementary angles ( $180^\circ$ ) at the centre.

Step 1 · Setup and Triangle Congruence

Let quadrilateral $ABCD$ circumscribe a circle with centre $O$ , touching sides $AB, BC, CD, DA$ at $P, Q, R, S$ respectively.

Join $OA, OB, OC, OD$ and radii $OP, OQ, OR, OS$ . Diagram 1

In $\Delta OAP$ and $\Delta OAS$ :

\begin{aligned} AP &= AS \quad (\text{Tangents from common external point } A) \\ OP &= OS \quad (\text{Radii of the same circle}) \\ OA &= OA \quad (\text{Common side}) \end{aligned}

By SSS congruence criterion: $\Delta OAP \cong \Delta OAS$

Step 2 · Establish Equal Angles at the Centre

Since corresponding parts of congruent triangles are equal (CPCT): $\angle POA = \angle SOA \implies \angle 1 = \angle 8$

Similarly, by proving the respective pairs of triangles congruent:

\begin{aligned} \Delta POB \cong \Delta QOB &\implies \angle 2 = \angle 3 \\ \Delta QOC \cong \Delta ROC &\implies \angle 4 = \angle 5 \\ \Delta ROD \cong \Delta SOD &\implies \angle 6 = \angle 7 \end{aligned}

Step 3 · Sum of Angles Around the Centre

The sum of all angles around the centre $O$ is $360^\circ$ : $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$

Substitute $\angle 8 = \angle 1$ , $\angle 3 = \angle 2$ , $\angle 4 = \angle 5$ , and $\angle 7 = \angle 6$ :

\begin{aligned} \angle 1 + \angle 2 + \angle 2 + \angle 5 + \angle 5 + \angle 6 + \angle 6 + \angle 1 &= 360^\circ \\ 2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 &= 360^\circ \\ 2(\angle 1 + \angle 2 + \angle 5 + \angle 6) &= 360^\circ \\ (\angle 1 + \angle 2) + (\angle 5 + \angle 6) &= 180^\circ \\ \angle AOB + \angle COD &= 180^\circ \end{aligned}

Similarly, substituting for the other pairs: $\angle BOC + \angle DOA = 180^\circ$

Answer

$\angle AOB + \angle COD = 180^\circ \quad \text{and} \quad \angle BOC + \angle DOA = 180^\circ$

Hence, opposite sides subtend supplementary angles at the centre.

Common Mistakes

Misgrouping Angles: Substituting the incorrect angle relations, which fails to combine adjacent angles into $\angle AOB$ and $\angle COD$ .
Confusing Cyclic with Circumscribed Quadrilaterals: Assuming the opposite interior angles of quadrilateral $ABCD$ sum to $180^\circ$ ; the theorem specifically applies to angles subtended at the centre of the circle by opposite sides.

Question 13

More questions in Exercise 10.2