Lesson Video: Proving Cyclic Quadrilaterals Mathematics

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

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Video Transcript

In this video, we will learn how to prove that a quadrilateral is cyclic by using the angles resulting from its diagonals. Let’s begin by defining a cyclic quadrilateral. A cyclic quadrilateral is a four-sided polygon whose vertices are inscribed on a circle. For example, this is a cyclic quadrilateral. 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.

Before we consider the properties of a cyclic quadrilateral, let’s recall two very important theorems about inscribed angles. An angle 𝜃 inscribed in a circle is half of the central angle two 𝜃 that subtends the same arc on the circle. In other words, the angle at the circumference is half the angle at the center. This then leads into a second inscribed angle theorem, which tells us that inscribed angles subtended by the same arc are equal. So let’s see how these will be useful when we look at the proof of cyclic quadrilaterals.

We can take the example of this cyclic quadrilateral 𝐴𝐵𝐶𝐷. And let’s draw in its diagonal line segments. Using the arc 𝐷𝐶 and given that the angles subtended by the same arc are equal, then we can say that the measure of angle 𝐷𝐴𝐶 is equal to the measure of angle 𝐷𝐵𝐶. We can then use the same property with this arc 𝐴𝐵 to show that the measure of angle 𝐴𝐷𝐵 must be equal to the measure of angle 𝐴𝐶𝐵.

We can then observe that, 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. In this example, we found two pairs of congruent angles. But we can also use the arc 𝐵𝐶 to show that the measure of angle 𝐵𝐴𝐶 is equal to the measure of angle 𝐵𝐷𝐶. Using the arc 𝐴𝐷 would show that the measure of angle 𝐴𝐵𝐷 is equal to the measure of angle 𝐴𝐶𝐷.

When it comes to proving if a quadrilateral is cyclic, we’ll need to see if the converse of this theorem is true. Let’s see if we can prove that if the angles created by the diagonals are equal, then that means that the quadrilateral is cyclic. Let’s take a different quadrilateral, 𝐴𝐵𝐶𝐷, along with its diagonals. If we can prove that the measure of angle 𝐷𝐴𝐶 is equal to the measure of angle 𝐷𝐵𝐶, then the quadrilateral is cyclic. This is because 𝐷𝐶 must be an arc of the circle. Therefore, 𝐴 and 𝐵 must also be points on the same circle. Therefore, every vertex must be on the circle. And this is by definition a cyclic quadrilateral.

Of course, it doesn’t always have to be the top two angles here that we proved that are congruent. For example, if we could prove that the measure of angle 𝐴𝐷𝐵 is equal to the measure of angle 𝐴𝐶𝐵, then this would also demonstrate that the quadrilateral is cyclic. However, we just need one of these pairs of congruent angles to demonstrate that the quadrilateral is cyclic. You might also wonder if perhaps every quadrilateral is cyclic. But let’s have a look at a different example.

Here is quadrilateral 𝐸𝐹𝐺𝐻. We can observe by eye that the measure of angle 𝐺𝐸𝐻 is not equal to the measure of angle 𝐻𝐹𝐺. We could not draw a circle that passes through all four vertices. And so 𝐸𝐹𝐺𝐻 is not a cyclic quadrilateral.

We’ll now look at some examples where we prove if a quadrilateral is cyclic or not.

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

If there is a circle passing through the vertices of this quadrilateral, then it would be a cyclic quadrilateral. There are a number of different angle properties we can use to determine if this quadrilateral is cyclic. However, given we have the diagonals marked, let’s check the angles made with the diagonals. We can then pose the question, is there an angle made with the diagonal and side which is equal in measure to the angle created by the other diagonal and opposite side?

At the minute, we don’t have any congruent pairs of angles in the diagram. However, we might observe that this angle 𝐶𝐴𝐵 is an angle created by a side and diagonal. Angle 𝐶𝐷𝐵 is the angle created by the other diagonal and the opposite side. So let’s see if it’s congruent to angle 𝐶𝐴𝐵. Let’s consider the triangle 𝐶𝐵𝐷 and recall that the interior angles in a triangle sum to 180 degrees. Therefore, we can say that the three angles in this triangle, 54 degrees plus 79 degrees plus the measure of angle 𝐶𝐷𝐵, must be equal to 180 degrees. We can simplify the left-hand side and then subtract 133 degrees from both sides, which gives us that the measure of angle 𝐶𝐷𝐵 is 47 degrees.

This means that we now have a pair of congruent angles. The measure of angle 𝐶𝐷𝐵 is equal to the measure of angle 𝐶𝐴𝐵. Therefore, we can say that an angle made with a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side. Therefore, we can give the answer yes. Since we’ve shown that this is a cyclic quadrilateral, we could draw a circle which passes through all four vertices of 𝐴𝐵𝐶𝐷.

Let’s look at another example.

Is 𝐴𝐵𝐶𝐷 a cyclic quadrilateral?

We can observe in this figure that we have the two diagonals marked, 𝐴𝐶 and 𝐵𝐷. If we can prove that an angle created by a diagonal and side 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. The angle 𝐵𝐷𝐴 is an angle created by a diagonal and side. The angle created by the other diagonal and opposite side is angle 𝐵𝐶𝐴. If these two angles are congruent, then the quadrilateral is cyclic.

Another pair of angles we could check would be angle 𝐷𝐴𝐶 and angle 𝐷𝐵𝐶. If we knew that just one pair of these angle measures were congruent, then that would be sufficient to show that the quadrilateral is cyclic. So let’s see if we can work out this angle measure of 𝐵𝐶𝐴.

We are given that this angle measure 𝐵𝐸𝐴 is 90 degrees. And we can remember that the angles on a straight line sum to 180 degrees. So that means that angle 𝐵𝐸𝐶 must also measure 90 degrees. We can now consider the triangle 𝐵𝐸𝐶 and remember that the angles in a triangle add up to 180 degrees. And so we can write that 63 degrees plus 90 degrees plus the measure of angle 𝐵𝐶𝐸 is 180 degrees. Therefore, the measure of angle 𝐵𝐶𝐸 is equal to 180 degrees subtract 153 degrees, which is 27 degrees. Therefore, these two angle measures created at the diagonals are not equal. And this means that 𝐴𝐵𝐶𝐷 is not a cyclic quadrilateral. Therefore, we can give the answer no.

At the start of this question, we did also say that we could check the angle measures of 𝐶𝐵𝐷 and 𝐶𝐴𝐷. Calculating that angle 𝐴𝐸𝐷 is also 90 degrees, we could then establish that angle 𝐶𝐴𝐷 is 52 degrees. But of course 63 degrees is not equal to 52 degrees, once again showing that 𝐴𝐵𝐶𝐷 is not a cyclic quadrilateral.

In the next question, we’ll check if a given trapezoid is a cyclic quadrilateral.

Is the trapezoid 𝐴𝐵𝐶𝐷 a cyclic quadrilateral?

Given that we have a trapezoid, we should have one pair of parallel sides. And they’re marked here. 𝐵𝐶 and 𝐴𝐷 are parallel. As we have a transversal 𝐵𝐷, we can work out that the angle 𝐶𝐵𝐷 is alternate to the angle 𝐴𝐷𝐵. It’s also 84 degrees. We can label the intersection of the diagonals as point 𝐸. And then we can take a closer look at triangle 𝐵𝐸𝐶.

We can work out the measure of this unknown angle 𝐵𝐶𝐸 by remembering that the interior angles in a triangle add up to 180 degrees. Therefore, 84 degrees plus 52 degrees plus the measure of angle 𝐵𝐶𝐸 is equal to 180 degrees. 84 degrees plus 52 degrees gives 136 degrees. Subtracting 136 degrees from both sides gives us that the measure of angle 𝐵𝐶𝐸 is 44 degrees. So how does this help us work out if 𝐴𝐵𝐶𝐷 is cyclic or not?

Well, let’s consider that we were given an angle created by a diagonal and side, angle 𝐴𝐷𝐵. We’ve just calculated the angle created by the other diagonal and opposite side. If these two angle measures are congruent, then the quadrilateral is cyclic. But of course 84 degrees is not equal to 44 degrees. And so the two angle measures are not congruent. Therefore, 𝐴𝐵𝐶𝐷 is not cyclic. We can therefore answer the question with no.

So far, we have seen specific examples of different quadrilaterals. However, in the next two examples, we’ll consider general statements about sets of quadrilaterals, beginning with determining if all rectangles are cyclic quadrilaterals or not.

All rectangles are cyclic quadrilaterals. (A) True or (B) false.

We can begin by recalling that a cyclic quadrilateral is a quadrilateral with all four vertices inscribed on a circle. One way we can prove that a quadrilateral is cyclic is by demonstrating that an angle created by a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side. So let’s consider the properties of a rectangle, which is defined as a quadrilateral with four angles of 90 degrees. In a rectangle, there are also two pairs of opposite congruent sides.

We know that the diagonals of a rectangle divide the rectangle into two congruent triangles. Together, both diagonals create four congruent triangles. That is, any triangle that’s created by two sides and a diagonal is congruent to any other triangle that’s also created by two sides and a diagonal. So, for this rectangle, which we can call 𝐽𝐾𝐿𝑀, we can say that triangle 𝐽𝐾𝐿 is congruent to triangle 𝐿𝑀𝐽. We can also say that this same triangle 𝐿𝑀𝐽 is congruent to triangle 𝑀𝐿𝐾.

Within this last pair of congruent triangles, we can also state that two corresponding angle measures are congruent since the measure of angle 𝑀𝐽𝐿 is equal to the measure of angle 𝐿𝐾𝑀. This means that an angle created by a diagonal and side is equal in measure to the angle created by the other diagonal and opposite side. And so this rectangle and indeed any rectangle must be a cyclic quadrilateral.

As an alternative method, we could also consider that the diagonals of a rectangle are equal in length and bisect each other. This means that there will be four equal-length line segments extending from the point of intersection. These can be considered as radii extending from the center of a circle. Either method will allow us to give the answer true, since all rectangles are cyclic quadrilaterals.

In this example, we established that all rectangles are cyclic. However, it’s important to note that squares, which are defined as quadrilaterals with all sides equal and all internal angles equal to 90 degrees, are a subset of rectangles. Therefore, all squares are also cyclic quadrilaterals.

We’ll now look at a different type of quadrilateral.

All isosceles trapezoids are cyclic quadrilaterals. (A) True or (B) false.

Let’s begin by recalling that a trapezoid is a quadrilateral with one pair of parallel sides. An isosceles trapezoid is a special type of trapezoid which has the additional property that the two nonparallel sides, which are sometimes called legs, are equal in length. So let’s draw an isosceles trapezoid. It has one pair of parallel sides, and the other two nonparallel sides are equal in length. We can then use one of the properties of the diagonals of an isosceles trapezoid. The diagonals of an isosceles trapezoid create two congruent triangles at the legs. They also create two similar triangles at the bases.

Be careful, however, this rule only applies to isosceles trapezoids. For example, let’s take this trapezoid which is not isosceles. The diagonals do not create two congruent triangles at the legs. But let’s return to the isosceles trapezoid and look at the angles. We can define the intersection of the diagonals as point 𝐸. And then because we have two congruent triangles, we can identify corresponding angles. The measure of angle 𝐷𝐴𝐸 must be equal to the measure of angle 𝐶𝐵𝐸. Furthermore, the measure of angle 𝐴𝐷𝐸 is equal to the measure of angle 𝐵𝐶𝐸. Either of these pairs of congruent angles would demonstrate that the measure of an angle created by a diagonal and side is equal to the angle created by the other diagonal and opposite side. This means that isosceles trapezoid 𝐴𝐵𝐶𝐷, and any isosceles trapezoid, is a cyclic quadrilateral.

Although not all trapezoids are cyclic quadrilaterals, all isosceles trapezoids are. So we can give the answer true. We can summarize the previous two examples by saying that although there are some quadrilaterals which may be shown to be cyclic, there are three types which will always be cyclic. They are squares, rectangles, and isosceles trapezoids.

Now, let’s summarize the key points of this video. We began by noting that a cyclic quadrilateral is a four-sided polygon whose vertices are inscribed on a circle. 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. The converse of this 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 side is equal in measure to the angle created by the other diagonal and opposite side. Finally, we also saw that all squares, rectangles, and isosceles trapezoids are cyclic quadrilaterals.

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