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I'm Up and Down, and Round and Round - Concyclicity of Points

Grade 9CBSE

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

🔑Concepts

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Concyclicity refers to a set of points that all lie on the same circle.

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Three non-collinear points: There is one and only one circle passing through three given non-collinear points. If three points are collinear, no circle can pass through all of them.

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A quadrilateral is called a cyclic quadrilateral if all its four vertices lie on a circle.

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Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is 180∘180^\circ.

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Converse Theorem: If the sum of a pair of opposite angles of a quadrilateral is 180∘180^\circ, then the quadrilateral is cyclic.

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Equal Angles Theorem: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points are concyclic.

📐Formulae

∠A+∠C=180∘ and ∠B+∠D=180∘\angle A + \angle C = 180^\circ \text{ and } \angle B + \angle D = 180^\circ

If ∠ACB=∠ADB and C,D are on the same side of AB, then A,B,C,D are concyclic.\text{If } \angle ACB = \angle ADB \text{ and } C, D \text{ are on the same side of } AB, \text{ then } A, B, C, D \text{ are concyclic.}

💡Examples

Problem 1:

In a quadrilateral ABCDABCD, ∠ABC=75∘\angle ABC = 75^\circ and ∠ADC=105∘\angle ADC = 105^\circ. Determine if the points A,B,C,A, B, C, and DD are concyclic.

Solution:

We are given ∠ABC=75∘\angle ABC = 75^\circ and ∠ADC=105∘\angle ADC = 105^\circ. These are opposite angles of the quadrilateral ABCDABCD. Sum of opposite angles = ∠ABC+∠ADC=75∘+105∘=180∘\angle ABC + \angle ADC = 75^\circ + 105^\circ = 180^\circ.

Explanation:

According to the property of cyclic quadrilaterals, if the sum of a pair of opposite angles is 180∘180^\circ, the quadrilateral is cyclic. Therefore, the points A,B,C,A, B, C, and DD lie on a circle and are concyclic.

Problem 2:

ABCDABCD is a cyclic quadrilateral whose diagonals intersect at a point EE. If ∠DBC=70∘\angle DBC = 70^\circ and ∠BAC=30∘\angle BAC = 30^\circ, find ∠BCD\angle BCD. Further, if AB=BCAB = BC, find ∠ECD\angle ECD.

Solution:

∠CAD=∠DBC=70∘ (Angles in the same segment)\angle CAD = \angle DBC = 70^\circ \text{ (Angles in the same segment)} ∠DAB=∠CAD+∠BAC=70∘+30∘=100∘\angle DAB = \angle CAD + \angle BAC = 70^\circ + 30^\circ = 100^\circ Since ABCDABCD is a cyclic quadrilateral, ∠BCD+∠DAB=180∘\angle BCD + \angle DAB = 180^\circ ∠BCD+100∘=180∘  ⟹  ∠BCD=80∘\angle BCD + 100^\circ = 180^\circ \implies \angle BCD = 80^\circ If AB=BCAB = BC, then in △ABC\triangle ABC, ∠BCA=∠BAC=30∘\angle BCA = \angle BAC = 30^\circ (Angles opposite to equal sides). ∠ECD=∠BCD−∠BCA=80∘−30∘=50∘\angle ECD = \angle BCD - \angle BCA = 80^\circ - 30^\circ = 50^\circ

Explanation:

We use the property that angles subtended by the same arc at the circumference are equal (Angles in the same segment). Then we use the supplementary property of opposite angles in a cyclic quadrilateral to find ∠BCD\angle BCD.

Problem 3:

Two circles intersect at two points BB and CC. Through BB, two line segments ABDABD and PBQPBQ are drawn to intersect the circles at A,DA, D and P,QP, Q respectively. Prove that ∠ACP=∠QCD\angle ACP = \angle QCD.

Solution:

In the first circle, ∠ACP=∠ABP\angle ACP = \angle ABP (Angles in the same segment subtended by arc APAP). In the second circle, ∠QCD=∠QBD\angle QCD = \angle QBD (Angles in the same segment subtended by arc QDQD). Now, ∠ABP=∠QBD\angle ABP = \angle QBD because they are vertically opposite angles formed by the intersection of lines ADAD and PQPQ at BB. Therefore, ∠ACP=∠QCD\angle ACP = \angle QCD.

Explanation:

By identifying that the points A,C,B,PA, C, B, P are concyclic on the first circle and Q,C,B,DQ, C, B, D are concyclic on the second circle, we can equate angles in the same segments and link them via vertically opposite angles.