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Circles - Determine cyclicity of points and solve cyclic quadrilateral properties

Grade 9CBSE

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

🔑Concepts

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A quadrilateral is called cyclic if all four vertices lie on a single circle. These vertices are called concyclic points.

A cyclic quadrilateral ABCD with all four vertices on the circumference of a circle.
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The sum of either pair of opposite angles of a cyclic quadrilateral is 180∘180^\circ. Conversely, if the sum of a pair of opposite angles of a quadrilateral is 180∘180^\circ, the quadrilateral is cyclic.

Diagram showing that opposite angles A and C in a cyclic quadrilateral sum to 180 degrees.
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The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Exterior angle property: Exterior angle at C is equal to the opposite interior angle A.
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If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, the four points are concyclic.

📐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.

Problem 3:

In the figure, ABCDABCD is a cyclic quadrilateral in which ACAC and BDBD are its diagonals. If ∠DBC=55∘\angle DBC = 55^\circ and ∠BAC=45∘\angle BAC = 45^\circ, find ∠BCD\angle BCD.

Cyclic quadrilateral ABCD with diagonals AC and BD intersecting. Angle DBC is 55 degrees and angle BAC is 45 degrees.

Solution:

  1. In the same segment, angles subtended by the same chord are equal. Therefore, ∠CAD=∠DBC=55∘\angle CAD = \angle DBC = 55^\circ.
  2. Now, ∠DAB=∠CAD+∠BAC=55∘+45∘=100∘\angle DAB = \angle CAD + \angle BAC = 55^\circ + 45^\circ = 100^\circ.
  3. Since ABCDABCD is a cyclic quadrilateral, the sum of opposite angles is 180∘180^\circ.
  4. ∠DAB+∠BCD=180∘\angle DAB + \angle BCD = 180^\circ.
  5. 100∘+∠BCD=180∘  ⟹  ∠BCD=80∘100^\circ + \angle BCD = 180^\circ \implies \angle BCD = 80^\circ.

Explanation:

We use the property that angles in the same segment of a circle are equal to find the full angle at A, then use the cyclic quadrilateral property of supplementary opposite angles.

Problem 4:

In a cyclic quadrilateral PQRSPQRS, the side PQPQ is parallel to SRSR. If ∠P=110∘\angle P = 110^\circ, find the measure of ∠Q,∠R\angle Q, \angle R, and ∠S\angle S.

Cyclic quadrilateral PQRS where PQ is parallel to SR and angle P is 110 degrees.

Solution:

  1. In cyclic quadrilateral PQRSPQRS, opposite angles sum to 180∘180^\circ. Therefore, ∠P+∠R=180∘\angle P + \angle R = 180^\circ.
  2. 110∘+∠R=180∘  ⟹  ∠R=70∘110^\circ + \angle R = 180^\circ \implies \angle R = 70^\circ.
  3. Since PQ∥SRPQ \parallel SR, the consecutive interior angles sum to 180∘180^\circ. Therefore, ∠P+∠S=180∘\angle P + \angle S = 180^\circ.
  4. 110∘+∠S=180∘  ⟹  ∠S=70∘110^\circ + \angle S = 180^\circ \implies \angle S = 70^\circ.
  5. Similarly, for the cyclic property, ∠Q+∠S=180∘\angle Q + \angle S = 180^\circ.
  6. ∠Q+70∘=180∘  ⟹  ∠Q=110∘\angle Q + 70^\circ = 180^\circ \implies \angle Q = 110^\circ.

Explanation:

Because the quadrilateral is cyclic, opposite angles are supplementary. Because PQ∥SRPQ \parallel SR, adjacent angles between the parallels are also supplementary. This reveals that the quadrilateral is an isosceles trapezium.