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.
