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Circles - Cyclic Quadrilaterals

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A cyclic quadrilateral is defined as a quadrilateral where all four vertices lie on the circumference of a single circle. Visually, if you draw a circle and pick any four points A,B,C,DA, B, C, D on its edge and connect them to form a closed four-sided figure, it is a cyclic quadrilateral.

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The most fundamental property of a cyclic quadrilateral is that the sum of either pair of opposite angles is 180∘180^\circ. This is also known as the supplementary property of opposite angles. For a cyclic quadrilateral ABCDABCD, this means ∠A+∠C=180∘\angle A + \angle C = 180^\circ and ∠B+∠D=180∘\angle B + \angle D = 180^\circ.

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The converse property states that if the sum of a pair of opposite angles of a quadrilateral is 180∘180^\circ, then the quadrilateral is cyclic. This is a common method used in proofs to show that four points are concyclic (lie on the same circle).

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The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. Visually, if you extend one side of the quadrilateral, say side ABAB to a point EE outside the circle, the angle ∠CBE\angle CBE formed outside is equal to the angle ∠ADC\angle ADC inside the quadrilateral at the opposite vertex.

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If a side of a cyclic quadrilateral is also a diameter of the circle, the angle subtended by that side at any point on the remaining part of the circle is 90∘90^\circ. This means if ADAD is a diameter, the angle ∠ABD\angle ABD or ∠ACD\angle ACD will always be a right angle.

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A cyclic parallelogram is always a rectangle. This is because in a parallelogram, opposite angles are equal (∠A=∠C\angle A = \angle C), and in a cyclic quadrilateral, opposite angles sum to 180∘180^\circ (∠A+∠C=180∘\angle A + \angle C = 180^\circ). Combining these, 2∠A=180∘2\angle A = 180^\circ, so ∠A=90∘\angle A = 90^\circ, which defines a rectangle.

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Angles in the same segment of a circle are equal. When looking at a cyclic quadrilateral with its diagonals drawn, any two angles subtended by the same side (arc) at the circumference are equal. For example, ∠ACB=∠ADB\angle ACB = \angle ADB because they are both subtended by arc ABAB.

📐Formulae

∠A+∠C=180∘\angle A + \angle C = 180^\circ

∠B+∠D=180∘\angle B + \angle D = 180^\circ

Exterior ∠=Interior Opposite ∠\text{Exterior } \angle = \text{Interior Opposite } \angle

If ∠A+∠C=180∘  ⟹  ABCD is cyclic\text{If } \angle A + \angle C = 180^\circ \implies ABCD \text{ is cyclic}

💡Examples

Problem 1:

In a cyclic quadrilateral ABCDABCD, if ∠A=(2x+4)∘\angle A = (2x + 4)^\circ and ∠C=(4x−64)∘\angle C = (4x - 64)^\circ, find the value of xx and the measure of ∠A\angle A and ∠C\angle C.

Solution:

Step 1: We know that in a cyclic quadrilateral, the sum of opposite angles is 180∘180^\circ. Therefore, ∠A+∠C=180∘\angle A + \angle C = 180^\circ. Step 2: Substitute the given expressions: (2x+4)+(4x−64)=180(2x + 4) + (4x - 64) = 180. Step 3: Combine like terms: 6x−60=1806x - 60 = 180. Step 4: Solve for xx: 6x=240  ⟹  x=406x = 240 \implies x = 40. Step 5: Calculate the angles: ∠A=2(40)+4=84∘\angle A = 2(40) + 4 = 84^\circ and ∠C=4(40)−64=160−64=96∘\angle C = 4(40) - 64 = 160 - 64 = 96^\circ.

Explanation:

This problem applies the property that opposite angles of a cyclic quadrilateral are supplementary. By setting up a linear equation based on the sum being 180∘180^\circ, we can solve for the unknown variable.

Problem 2:

In the given figure of a cyclic quadrilateral ABCDABCD, side ABAB is produced to EE. If the exterior angle ∠CBE=85∘\angle CBE = 85^\circ and ∠CAD=40∘\angle CAD = 40^\circ, find ∠ACD\angle ACD given that CD=ADCD = AD.

Solution:

Step 1: Use the exterior angle property. The exterior angle ∠CBE\angle CBE is equal to the interior opposite angle ∠ADC\angle ADC. Thus, ∠ADC=85∘\angle ADC = 85^\circ. Step 2: In △ACD\triangle ACD, we are given CD=ADCD = AD. This means △ACD\triangle ACD is an isosceles triangle. Step 3: In an isosceles triangle, angles opposite to equal sides are equal. Therefore, ∠ACD=∠CAD\angle ACD = \angle CAD. Step 4: Since ∠CAD=40∘\angle CAD = 40^\circ, it follows that ∠ACD=40∘\angle ACD = 40^\circ.

Explanation:

This problem demonstrates the Exterior Angle Property of cyclic quadrilaterals and combines it with properties of isosceles triangles. The exterior angle helps identify one interior angle, which then allows us to use triangle properties to find others.