Circumscribing Quadrilaterals and The Rhombus Theorem for CBSE Class 10

When a polygon circumscribes a circle, every single one of its sides acts as a tangent touching the inscribed circle at a point of contact. This tight geometric constraint binds the side lengths and internal angles of the polygon into precise algebraic symmetries.

In CBSE Class 10 Mathematics, Chapter 10 (Circles), questions involving quadrilaterals circumscribing a circle appear in board examinations almost every single year. The benchmark identity $AB + CD = AD + BC$ and its corollary—proving that a circumscribing parallelogram is a rhombus—are two of the most celebrated 3-mark and 4-mark questions in the entire Class 10 syllabus.

What You Will Learn

• What does it mean for a quadrilateral to circumscribe a circle?
• Rigorous geometric proof of Theorem 1: $AB + CD = AD + BC$
• Rigorous geometric proof of Theorem 2: A parallelogram circumscribing a circle is a rhombus
• Rigorous geometric proof of Theorem 3: Opposite sides of a circumscribed quadrilateral subtend supplementary angles at the center
• Step-by-step solved CBSE board examination problems
• Methodical presentation templates and algebraic grouping tips

1. What is a Circumscribed Quadrilateral?

A quadrilateral $ABCD$ is said to circumscribe a circle (or the circle is inscribed in the quadrilateral) if all four sides of the quadrilateral ($AB, BC, CD,$ and $DA$) are tangents touching the circle at four distinct points of contact ($P, Q, R,$ and $S$).

                         D ---------- R ---------- C
                        /                                                  /                                                   S               O               Q
                       \                             /
                        \                           /
                         A ---------- P ---------- B

2. Theorem 1: Sum of Opposite Sides are Equal

Theorem Statement (CBSE Board Classic)

If a quadrilateral $ABCD$ is drawn to circumscribe a circle, then the sum of lengths of opposite sides is equal: $\mathbf{AB + CD = AD + BC}$

Given:

A quadrilateral $ABCD$ circumscribes a circle with center $O$, touching sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively.

To Prove:

$AB + CD = AD + BC$

Step-by-Step Geometric Proof:

1. Recall Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
2. Set up Equations from Each Vertex:
   • From external vertex $A$: $AP = AS \quad \text{--- (1)}$
   • From external vertex $B$: $BP = BQ \quad \text{--- (2)}$
   • From external vertex $C$: $CR = CQ \quad \text{--- (3)}$
   • From external vertex $D$: $DR = DS \quad \text{--- (4)}$
3. Add Equations (1), (2), (3), and (4) together: $(AP + BP) + (CR + DR) = (AS + BQ + CQ + DS)$
4. Regroup the terms to form complete sides:
   • On LHS: $(AP + BP) = AB$ and $(CR + DR) = CD$.
   • On RHS: Group $(AS + DS)$ together, and group $(BQ + CQ)$ together: $(AS + DS) = AD \quad \text{and} \quad (BQ + CQ) = BC$
5. Combine both sides: $\mathbf{AB + CD = AD + BC}$ Hence, proved.

Important: <u>Notice how we aligned the equations! In Equations (1) to (4), make sure that terms forming the same side (like $AP$ and $BP$, or $CR$ and $DR$) are placed on the SAME side of the equal sign. If you write $BQ = BP$, then adding will mix sides up!</u>

3. Theorem 2: Parallelogram Circumscribing a Circle is a Rhombus

Theorem Statement (NCERT & CBSE Board Classic)

Prove that the parallelogram circumscribing a circle is a rhombus.

Given:

A parallelogram $ABCD$ circumscribes a circle.

To Prove:

$ABCD$ is a rhombus (all four sides are equal: $AB = BC = CD = DA$).

Step-by-Step Proof:

1. Since $ABCD$ is a parallelogram, its opposite sides are equal: $AB = CD \quad \text{--- (1)}$ $AD = BC \quad \text{--- (2)}$
2. Since parallelogram $ABCD$ circumscribes a circle, by Theorem 1: $AB + CD = AD + BC \quad \text{--- (3)}$
3. Substitute $CD = AB$ and $AD = BC$ into Equation (3): $AB + AB = BC + BC$ $2AB = 2BC$
4. Divide both sides by 2: $\mathbf{AB = BC}$
5. Combine with parallelogram properties: From (1) and (2), since $AB = CD$ and $BC = AD$, and now $AB = BC$: $\mathbf{AB = BC = CD = DA}$
6. Since all four sides of parallelogram $ABCD$ are equal, <u>quadrilateral $ABCD$ is a rhombus</u>. Hence, proved.

4. Theorem 3: Supplementary Angles Subtended at the Center

Theorem Statement (CBSE 4-Mark Board Question)

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

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

                                      D ---- R ---- C
                                     / \   4/3   /                                     /   \  / \  /                                      S   5 \/   \/ 2   Q
                                   |     /\ O /\     |
                                   \   6/  \ /  
  /
                                    \  /   7\8   \ /
                                     A ----- P --- B

Outline of Proof:

1. Join center $O$ to all four vertices ($A, B, C, D$) and to all four contact points ($P, Q, R, S$).
2. This creates 8 small triangles around center $O$.
3. By RHS congruence:
   • $\Delta OAP \cong \Delta OAS \implies \angle 8 = \angle 7$
   • $\Delta OBP \cong \Delta OBQ \implies \angle 1 = \angle 2$
   • $\Delta OCQ \cong \Delta OCR \implies \angle 3 = \angle 4$
   • $\Delta ODR \cong \Delta ODS \implies \angle 5 = \angle 6$
4. The sum of all angles around point $O$ is $360^\circ$: $(\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8) = 360^\circ$
5. Group pairs: $2\angle 1 + 2\angle 8 + 2\angle 4 + 2\angle 5 = 360^\circ$ $2(\angle 1 + \angle 8) + 2(\angle 4 + \angle 5) = 360^\circ$
6. Notice that $(\angle 1 + \angle 8) = \angle AOB$ and $(\angle 4 + \angle 5) = \angle COD$: $2\angle AOB + 2\angle COD = 360^\circ$ $\mathbf{\angle AOB + \angle COD = 180^\circ}$
7. Similarly, $\mathbf{\angle BOC + \angle AOD = 180^\circ}$. Hence, proved.

5. Summary and Examination Tips

Exam Tip: In the rhombus proof, always prove the identity $AB + CD = AD + BC$ first before applying the parallelogram condition! Jumping straight to $2AB = 2BC$ without proving the addition of tangents will lose you 2 marks.

Common Mistake: Mixing up adjacent and opposite sides. The theorem states that the sum of opposite sides are equal ($AB + CD = AD + BC$), NOT adjacent sides!