Lesson Explainer: Proving Cyclic Quadrilaterals Mathematics

In this explainer, we will learn how to prove that a quadrilateral is cyclic using the angles resulting from its diagonals.

We can begin by recapping what is meant by an inscribed angle.

Definition: Inscribed Angle

An inscribed angle is the angle made when two chords intersect on the circumference of the circle. The vertex of the angle lies on the circumference of the circle.

We use our understanding of inscribed angles to define a cyclic quadrilateral.

Definition: Cyclic Quadrilateral

A cyclic quadrilateral is a four-sided polygon whose vertices lie on a circle, so all its angles are inscribed angles in the circle.

Before we consider the properties of a cyclic quadrilateral, we recap two important theorems about inscribed angles and central angles (an angle at the center of a circle with endpoints on its circumference).

Definition: Inscribed Angle Theorems

An angle, πœƒ, inscribed in a circle is half of the central angle, 2πœƒ, that is subtended by the same arc on the circle. In other words, the angle at the circumference is half the angle at the center.

Inscribed angles subtended by the same arc are equal.

We can use these properties to investigate the properties of a cyclic quadrilateral. Let’s consider the cyclic quadrilateral 𝐴𝐡𝐢𝐷 with its diagonal line segments.

Since we have an arc 𝐷𝐢, then by using the fact that inscribed angles subtended by the same arc are equal, we can state that π‘šβˆ π·π΄πΆ=π‘šβˆ π·π΅πΆ.

We could also observe by using the same property with the arc 𝐴𝐡 that π‘šβˆ π΄π·π΅=π‘šβˆ π΄πΆπ΅.

So, in any cyclic quadrilateral, the angle created by a diagonal and side is equal to the angle created by the other diagonal and opposite side.

Property: Angle Measures between the Diagonals and Sides of a Cyclic Quadrilateral

In a cyclic quadrilateral, the angle created by a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side. This pair of angles consists of two inscribed angles that are subtended by the same arc.

The converse of this theorem is also true. That is, in a given quadrilateral, if we can prove that the angles created by the diagonals are equal, then the quadrilateral is cyclic.

For example, given the following quadrilateral, 𝐴𝐡𝐢𝐷, if we can prove that π‘šβˆ π·π΄πΆ=π‘šβˆ π·π΅πΆ or π‘šβˆ π΄π·π΅=π‘šβˆ π΄πΆπ΅, then the quadrilateral is cyclic. That is, a circle can be constructed through all four of its vertices.

Note that we do not have to prove both of these pairs of angles are equal. If we just have π‘šβˆ π·π΄πΆ=π‘šβˆ π·π΅πΆ, then we know that 𝐷𝐢 must be an arc and 𝐴 and 𝐡 are points on the same circle. Therefore, every point lies on the same circle and it is, by definition, a cyclic quadrilateral.

We can see how in a noncyclic quadrilateral, this property would not hold true. In the quadrilateral below, we can observe by eye that π‘šβˆ πΊπΈπ»β‰ π‘šβˆ π»πΉπΊ.

We could not draw a circle passing through all 4 vertices, and so, this quadrilateral is not cyclic.

We will see how we can apply this rule to identify cyclic quadrilaterals in the following examples.

Example 1: Determining Whether a Given Quadrilateral Is a Cyclic Quadrilateral

Is there a circle passing through the vertices of the quadrilateral 𝐴𝐡𝐢𝐷?

Answer

A quadrilateral that has all four vertices inscribed in a circle is defined as a cyclic quadrilateral. We can use the inscribed angle properties to determine if a quadrilateral is cyclic, and given that we have the diagonals drawn, we can check if the angle made with a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side.

Let’s see if we can calculate π‘šβˆ πΆπ·π΅ using △𝐢𝐷𝐡. We recall that the interior angles in a triangle sum to 180∘. Given that π‘šβˆ π·π΅πΆ=54∘ and π‘šβˆ π΅πΆπ·=79∘, we have π‘šβˆ π·π΅πΆ+π‘šβˆ π΅πΆπ·+π‘šβˆ πΆπ·π΅=18054+79+π‘šβˆ πΆπ·π΅=180133+π‘šβˆ πΆπ·π΅=180π‘šβˆ πΆπ·π΅=47.∘∘∘∘∘∘∘

We now have a pair of congruent angles, π‘šβˆ πΆπ·π΅=π‘šβˆ πΆπ΄π΅. Thus, we have demonstrated that the angle made with a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side, and so, 𝐴𝐡𝐢𝐷 is a cyclic quadrilateral. Alternatively, we can consider that this property arises from the fact that 𝐡 and 𝐢 form an arc, and since the angles at 𝐷 and 𝐴 are equal, then they both lie on the circumference of the same circle.

We could even draw the circle through all four vertices if we wished.

Thus, we have the answer: yes, since 𝐴𝐡𝐢𝐷 is a cyclic quadrilateral, we could create a circle passing through all the vertices of 𝐴𝐡𝐢𝐷.

In the next example, we will see an example of how if we can demonstrate that the angles at the diagonals are not equal, then the quadrilateral is not cyclic.

Example 2: Determining Whether a Given Quadrilateral Is a Cyclic Quadrilateral

Is 𝐴𝐡𝐢𝐷 a cyclic quadrilateral?

Answer

We can observe that, in the figure, the diagonals are marked, along with two given angle measures. If we can demonstrate that the angle created with a diagonal and a side of the quadrilateral is equal in measure to the angle created by the other diagonal and opposite side, then the quadrilateral is cyclic. If they are not equal, then it is not a cyclic quadrilateral.

Given that we have π‘šβˆ π΅π·π΄=38∘, then we could attempt to calculate π‘šβˆ π΅πΆπ΄. Alternatively, another potential angle pair to compare would be ∠𝐷𝐴𝐢 and ∠𝐷𝐡𝐢. Knowing both values of one pair of these angles would be sufficient to prove if the quadrilateral is cyclic or not.

Let’s consider the first pair of angles and seek to calculate the measure of ∠𝐡𝐢𝐴. We can define the intersection point of the diagonals as 𝐸. Note that we are given that π‘šβˆ π΅πΈπ΄=90∘ and we recall that the angles on a straight line sum to 180∘. Thus, we have π‘šβˆ π΅πΈπΆ=180βˆ’π‘šβˆ π΅πΈπ΄=180βˆ’90=90.∘∘∘∘

We now have △𝐡𝐢𝐸 with two known angles, since we are given that π‘šβˆ πΆπ΅πΈ=63∘, and so, we can calculate the third angle, π‘šβˆ π΅πΆπΈ, using the property that the sum of the internal angles in a triangle is 180∘. Hence, π‘šβˆ πΆπ΅πΈ+π‘šβˆ π΅πΈπΆ+π‘šβˆ π΅πΆπΈ=18063+90+π‘šβˆ π΅πΆπΈ=180153+π‘šβˆ π΅πΆπΈ=180π‘šβˆ π΅πΆπΈ=27.∘∘∘∘∘∘∘

Checking the angles created at the diagonals, we observe that 27β‰ 38,∘∘ so π‘šβˆ π΅πΆπ΄β‰ π‘šβˆ π΅π·π΄.

Therefore, the answer is no, 𝐴𝐡𝐢𝐷 is not a cyclic quadrilateral.

We could have alternatively calculated that π‘šβˆ π·π΄πΆ=52∘.

Since this angle is not equal to π‘šβˆ π·π΅πΆ(63)∘, then this would also have demonstrated that the angles created by the diagonals are not equal, and thus, the quadrilateral 𝐴𝐡𝐢𝐷 is not cyclic.

Let’s now look at an example involving a trapezoid.

Example 3: Determining Whether a Given Quadrilateral Is a Cyclic Quadrilateral

Is the trapezoid 𝐴𝐡𝐢𝐷 a cyclic quadrilateral?

Answer

We can observe that there are two parallel sides in 𝐴𝐡𝐢𝐷, since 𝐡𝐢⫽𝐴𝐷. Given these parallel lines and the transversal 𝐡𝐷, we can observe that ∠𝐷𝐡𝐢 is alternate to the given angle 𝐴𝐷𝐡. These two angles will be equal in measure. Hence, π‘šβˆ π·π΅πΆ=π‘šβˆ π΄π·π΅=84.∘

Next, we can use the fact that the sum of the interior angles in a triangle is 180∘. We can define the intersection of the diagonals as point 𝐸.

Thus, in triangle 𝐡𝐢𝐸, we calculate π‘šβˆ πΈπ΅πΆ+π‘šβˆ π΅πΈπΆ+π‘šβˆ π΅πΆπΈ=18084+52+π‘šβˆ π΅πΆπΈ=180136+π‘šβˆ π΅πΆπΈ=180π‘šβˆ π΅πΆπΈ=44.∘∘∘∘∘∘∘

We now have enough information to determine whether 𝐴𝐡𝐢𝐷 is a cyclic quadrilateral. We have the measures of two angles each created by a diagonal and side, ∠𝐡𝐢𝐸 and ∠𝐴𝐷𝐸. If these two angles are equal, then the quadrilateral is cyclic.

However, since π‘šβˆ π΅πΆπΈ(=44)β‰ π‘šβˆ π΄π·πΈ(=84),∘∘ then all four vertices cannot lie on a circle.

Therefore, we have the answer: no, 𝐴𝐡𝐢𝐷 is not a cyclic quadrilateral.

Let’s look at another example.

Example 4: Using the Properties of Cyclic Quadrilaterals to Verify Whether the Given Quadrilateral Is Cyclic

Is 𝐴𝐡𝐢𝐷 a cyclic quadrilateral?

Answer

We can begin by recalling that a cyclic quadrilateral is a quadrilateral with all four vertices lying on a circle. One way we can prove a quadrilateral is cyclic is by demonstrating that an angle created by a diagonal and a side is equal in measure to the angle created by the other diagonal and the opposite side.

Using the property that the angle measures on a straight line sum to 180∘, we can calculate π‘šβˆ π·πΈπ΄ as π‘šβˆ π·πΈπ΄+π‘šβˆ π·πΈπΆ=180π‘šβˆ π·πΈπ΄+83=180π‘šβˆ π·πΈπ΄=180βˆ’83=97.∘∘∘∘∘∘

Next, we can use the given angle measure, π‘šβˆ π·π΄πΈ=42∘, and the property that the sum of the angle measures in a triangle is 180∘ to calculate π‘šβˆ π΄π·πΈ. Thus, we have π‘šβˆ π΄π·πΈ+π‘šβˆ π·π΄πΈ+π‘šβˆ π·πΈπ΄=180π‘šβˆ π΄π·πΈ+42+97=180π‘šβˆ π΄π·πΈ+139=180π‘šβˆ π΄π·πΈ=180βˆ’139=41.∘∘∘∘∘∘∘∘∘

We can now observe that π‘šβˆ π΄πΆπ΅(41)=π‘šβˆ π΄π·π΅(41).∘∘

Hence, we can give the answer: yes, 𝐴𝐡𝐢𝐷 is a cyclic quadrilateral.

In the previous examples, we have seen specific examples of different quadrilaterals. In the next two examples, we will consider general statements about a set of quadrilaterals, beginning with determining if all rhombuses are cyclic quadrilaterals.

Example 5: Determining Whether a Rhombus Is a Cyclic Quadrilateral

True or False: All rhombuses are cyclic quadrilaterals.

  1. True
  2. False

Answer

We can begin by recalling that a cyclic quadrilateral is a quadrilateral with all four vertices inscribed in a circle. One way we can prove a quadrilateral is cyclic is by demonstrating that an angle created by a diagonal and a side is equal in measure to the angle created by the other diagonal and the opposite side.

A rhombus is defined as a quadrilateral with all four sides equal in length. A rhombus has the properties that there are two pairs of parallel sides, and each of the two diagonals is the perpendicular bisector of the other. We could sketch a rhombus 𝐴𝐡𝐢𝐷 below.

Let’s consider one of the interior angles, ∠𝐷𝐴𝐢. If the rhombus is cyclic, then the angle 𝐷𝐡𝐢 would be congruent. However, the only angle that we can prove is congruent to ∠𝐷𝐴𝐢 is ∠𝐴𝐢𝐡, since 𝐴𝐷 and 𝐡𝐢 are parallel lines that are intersected by a transversal, 𝐴𝐢.

Similarly, using the properties of parallel lines, ∠𝐴𝐷𝐡 is congruent to ∠𝐷𝐡𝐢.

We can observe simply by eye that, in this rhombus, ∠𝐷𝐴𝐢 is not congruent to ∠𝐷𝐡𝐢. In fact, the only way in which these two angles would be congruent is if all four marked angles (∠𝐷𝐴𝐢, ∠𝐴𝐢𝐡, ∠𝐴𝐷𝐡, and ∠𝐷𝐡𝐢) are congruent, in which case the rhombus would have 4 interior angles of 90∘ and would be a square.

Therefore, we have demonstrated that although a rhombus may be a cyclic quadrilateral if it is a square, which is a special case of a rhombus, we cannot say that all rhombuses are cyclic quadrilaterals. Hence, the statement that all rhombuses are cyclic quadrilaterals is false.

In the previous example, we established that a rhombus is only a cyclic quadrilateral in the special case where the rhombus is a square. All squares are cyclic quadrilaterals. All rectangles are also cyclic quadrilaterals.

In the final example, we consider if all isosceles trapezoids are cyclic quadrilaterals.

Example 6: Determining Whether an Isosceles Trapezoid Is a Cyclic Quadrilateral

True or False: All isosceles trapezoids are cyclic quadrilaterals.

  1. True
  2. False

Answer

Let’s recall that a trapezoid is a quadrilateral with one pair of parallel sides. An isosceles trapezoid is a special type of trapezoid, with the additional property that the two nonparallel sides (the β€œlegs”) are equal in length.

We can sketch the diagonals of an isosceles trapezoid, and we recall that, in any isosceles trapezoid, the two angles at any base of the two parallel bases are equal in measure, so at the base 𝐢𝐷, we have π‘šβˆ π΄π·πΆ=π‘šβˆ π΅πΆπ·.

We can also observe that in the triangles 𝐡𝐢𝐷 and 𝐴𝐷𝐢, we have two pairs of congruent sides. The legs of the trapezoid are congruent, so 𝐡𝐢=𝐴𝐷 and they have a shared side. 𝐢𝐷 is a common side in both triangles.

Hence, △𝐡𝐢𝐷≅△𝐴𝐷𝐢, according to the side-angle-side (SAS) criteria. So, we can deduce that π‘šβˆ πΆπ΅π·=π‘šβˆ π·π΄πΆ.

Thus, we have that the measure of an angle created by a diagonal and a side, ∠𝐢𝐡𝐷, is equal to the measure of the angle created by the other diagonal and opposite side, ∠𝐷𝐴𝐢. This means that the isosceles trapezoid 𝐴𝐡𝐢𝐷, and any isosceles trapezoid in general, is cyclic.

Although not all trapezoids are cyclic, all isosceles trapezoids are, so the statement in the question is true.

We have demonstrated that all isosceles trapezoids are cyclic quadrilaterals but not all trapezoids. We saw an example of this in the third example in this explainer. The given trapezoid in this problem was shown to be noncyclic.

Any type of quadrilateral may be proven to be cyclic; however, there are just 3 types that are always cyclic: squares, rectangles, and isosceles trapezoids.

We will now summarize the key points of this explainer.

Key Points

  • A cyclic quadrilateral is a four-sided polygon whose vertices lie on a circle, so all its angles are inscribed angles in the circle.
  • In a cyclic quadrilateral, the angle created by a diagonal and a side is equal in measure to the angle created by the other diagonal and the opposite side, because they are subtended by the same arc in the circle.
  • The converse of the above is also true. Thus, one method we can use to prove a quadrilateral is cyclic is by demonstrating that an angle created by a diagonal and a side is equal in measure to the angle created by the other diagonal and the opposite side.
  • All squares, rectangles, and isosceles trapezoids are cyclic quadrilaterals.

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