Video Transcript
In this video, we’ll learn how to
use the theorems of intersecting chords, secants, or tangents and secants to find
missing lengths in a circle.
Let’s begin by recapping the names
of the various parts of a circle. By this stage, you should feel
confident in being able to identify a chord, a radius, or a diameter of a
circle. Then we know that a line that
intersects the circumference of a circle exactly once and is perpendicular to the
diameter of that circle is called the tangent. If the tangent has an endpoint on
the circumference of the circle, we then call it the tangent segment. And it’s important to realize that
the tangent segment extends in one direction infinitely.
Next, we have a secant line. Now that’s a line that intersects a
curve at a minimum of two distinct points. The secant to a circle intersects
the circumference exactly twice, whilst a secant segment will intersect twice, but
an endpoint will lie on the circumference of that circle. So now we have these definitions;
we’re going to consider a couple of theorems that can help us to solve problems
involving circles.
The first is the intersecting
chords theorem. So we have a pair of chords 𝐴𝐵
and 𝐶𝐷 that intersect at the point 𝐸. The intersecting chords theorem
tells us that the products of the length of the line segments on each chord are
equal. So in this case, 𝐴𝐸 times 𝐸𝐵 is
equal to the product of 𝐶𝐸 and 𝐸𝐷. If we define line segment 𝐴𝐸 to
be 𝑎 units, 𝐸𝐵 to be 𝑏 units, and so one, we can alternatively write this as 𝑎
times 𝑏 equals 𝑐 times 𝑑. And then this is really useful
because we can rearrange this to represent these lengths proportionally such that 𝑎
over 𝑐 is equal to 𝑑 over 𝑏. And so the benefit of this theorem,
is that if we know any of these three values, we can find the fourth.
Now that we have the intersecting
chords theorem, we’re going to implement this in an example to help us find a
missing length.
If 𝐸𝐴 over 𝐸𝐵 is equal to
eight-sevenths, 𝐸𝐶 is equal to seven centimeters, and 𝐸𝐷 is equal to eight
centimeters, find the length of line segments 𝐸𝐵 and 𝐸𝐴.
And we also have a diagram of a
circle which has two chords, 𝐴𝐵 and 𝐶𝐷. We notice that these chords
intersect at a point. That’s point 𝐸. And this is really useful because
we can use the intersecting chords theorem to link the lengths of line segments
𝐸𝐶, 𝐸𝐵, 𝐸𝐴, and 𝐸𝐷. The intersecting chords theorem
tells us that the product of 𝐸𝐶 and 𝐸𝐷 is equal to the product of the lengths of
line segments 𝐸𝐵 and 𝐸𝐴. In this case then, 𝐸𝐶 times 𝐸𝐷
is a 𝐸𝐵 times 𝐸𝐴.
The question tells us that 𝐸𝐶 is
equal to seven centimeters whilst 𝐸𝐷 is equal to eight centimeters. Substituting these values into our
formula then, we get seven times eight is equal to 𝐸𝐵 times 𝐸𝐴 or 56 is equal to
𝐸𝐵 times 𝐸𝐴. And at first glance, it might look
like we don’t have enough information to answer this question. But we haven’t used the
proportional relationship between 𝐸𝐴 and 𝐸𝐵. That is, 𝐸𝐴 over 𝐸𝐵 is equal to
eight-sevenths. By multiplying both sides of this
equation by 𝐸𝐵, we find that 𝐸𝐴 is equal to eight-sevenths 𝐸𝐵. And then we can replace 𝐸𝐴 with
this expression in our earlier equation.
This then becomes 56 is equal to
𝐸𝐵 times eight-sevenths 𝐸𝐵 or 56 is eight-sevenths 𝐸𝐵 squared. Let’s solve this equation by
dividing by eight-sevenths. Now that’s the same as dividing by
eight and then multiplying by seven, giving us 49 is equal to 𝐸𝐵 squared. The final step is to then find the
positive square root of 49. Now, usually we would find both the
positive and negative square root, but of course here this is giving us a
length. So we’re only interested in the
positive value. 𝐸𝐵 is therefore equal to seven or
seven centimeters.
Once we have this value, we can use
our earlier equation, that is, 𝐸𝐴 equals eight-sevenths 𝐸𝐵, to find the value of
𝐸𝐴. 𝐸𝐴 is then eight-sevenths times
seven. Well, that, of course, is simply
equal to eight or eight centimeters. The length of line segment 𝐸𝐵 is
seven centimeters, and the length of line segment 𝐸𝐴 is eight centimeters.
So we’ve demonstrated the
intersecting chords theorem. We’re now going to introduce a
second theorem that’s going to help us solve missing value problems with
circles. This theorem states that the
product of the measures of one secant segment and its external secant segment is
equal to the product of the measures of the other secant segment and its external
secant segment, where the two secant segments, of course, must intersect. In our diagram, that is 𝐴𝐸 times
𝐵𝐸 equals 𝐶𝐸 times 𝐷𝐸.
Labeling the various segments as 𝑎
units, 𝑏 units, and so on, we can alternatively write this as 𝑎 times 𝑏 equals 𝑐
times 𝑑. And this is useful to do because
the intersecting secants theorem has a special case. And we call it the tangent secant
theorem. If one or both of the lines are
tangent segments, then according to our diagram here we can say that 𝐴𝐵 times 𝐵𝐸
is equal to the square of 𝐷𝐸. So 𝑎 times 𝑏 is 𝑑 squared. We’ll now demonstrate how to use
one of these theorems to solve a problem involving two secant lines that intersect
outside of the circle.
If 𝐸𝐶 is equal to 10 centimeters,
𝐸𝐷 is equal to six centimeters, 𝐸𝐵 equals five centimeters, find the length of
line segment 𝐸𝐴.
Then we have a diagram with a
circle with two intersecting secants drawn on it. That’s 𝐴𝐸 and 𝐶𝐸. Since we’re dealing with a pair of
intersecting secant segments, we’ll use the intersecting secants theorem. This is that the product of the
measures of one secant segment and its external secant segment is equal to the
product of the measures of the other secant segment and its external secant
segment. So in the case of our diagram,
that’s 𝐸𝐴 times 𝐸𝐵 equals 𝐸𝐷 times 𝐸𝐶.
Now we’re told in the question the
length of line segment 𝐸𝐶 to be 10 centimeters, 𝐸𝐷 is six centimeters, and 𝐸𝐵
is five centimeters. So let’s substitute these
dimensions into our equation. When we do, we find that 𝐸𝐴 times
five is equal to six times 10. Then we solve this equation for
𝐸𝐴 by dividing both sides by five. So 𝐸𝐴 is six times 10 divided by
five. Now we could evaluate the numerator
of this fraction. Or alternatively, we spot that
there is a common factor between 10 and five. We can divide them both by
five. This means that 𝐸𝐴 is equal to
six times two divided by one. And that’s 12 or 12
centimeters. The length of line segment 𝐸𝐴 is
then 12 centimeters.
In our next example, we’ll
demonstrate how to use the special case of the intersecting secants theorem, that
is, the tangent secant theorem, where one of the lengths is, in fact, a tangent
segment.
In the figure shown, the circle has
a radius of 12 centimeters. 𝐴𝐵 is equal to 12 centimeters,
and 𝐴𝐶 is equal to 35 centimeters. Determine the distance from line
segment 𝐵𝐶 to the center of the circle 𝑀 and the length of line segment 𝐴𝐷,
rounding your answers to the nearest tenth.
We’re going to begin by finding the
distance from line segment 𝐵𝐶 to the center of the circle 𝑀. We might recall that the shortest
distance from a point to a line is the length of the perpendicular from that point
to the line. And so we construct this
perpendicular from point 𝑀 to the line segment 𝐵𝐶. In fact, since 𝑀 is the center of
the circle and 𝐵𝐶 is a chord, we can say that this perpendicular is the
perpendicular line bisector of 𝐵𝐶. So defining the point where this
perpendicular meets the line segment 𝐵𝐶 as 𝐸, we can say that 𝐵𝐸 must be equal
to 𝐸𝐶.
Next, we’re going to use the fact
that the radius of the circle is 12 centimeters. The radius, of course, is the line
segment that joins the point of the center of the circle to any point on its the
circumference. So we can say that 𝑀𝐵 is 12
centimeters.
Next, we apply the fact that 𝐴𝐵
is equal to 12 centimeters and 𝐴𝐶 is equal to 35 centimeters. Since we can think of line segment
𝐴𝐶 as the sum of line segments 𝐴𝐵 and 𝐵𝐶, we can say that 35 is equal to 12
plus 𝐵𝐶 and we can find the length of 𝐵𝐶 by subtracting 12 from both sides of
this equation. 35 minus 12 is 23. So 𝐵𝐶 is 23 centimeters in
length.
But remember, we said that the line
segment 𝑀𝐸 is the perpendicular bisector for the line segment 𝐵𝐶. So 𝐵𝐸 must be half of 𝐵𝐶, that
is, 23 divided by two or 23 over two centimeters. We now note that we have a right
triangle 𝑀𝐸𝐵 for which we know two of its sides. We can therefore use the
Pythagorean theorem to find the length of the side 𝑀𝐸. Let’s call that 𝑥 or 𝑥
centimeters.
Substituting what we know about
this triangle into the Pythagorean theorem, and we find that 12 squared equals 𝑥
squared plus 23 over two squared. Then we make 𝑥 squared the subject
by subtracting 23 over two squared from both sides. 12 squared minus 23 over two
squared is 47 over four. To find the length that we’re
interested in, 𝑥, we’re going to find the positive square root of 47 over four. And that’s equal to 3.427 and so
on. Correct the nearest tenth, we find
that’s equal to 3.4 centimeters.
We now move on to the second part
of this question. And that asks us to find the length
of line segment 𝐴𝐷. And we observed that line segment
𝐴𝐷 is, in fact, a tangent segment, whilst the line 𝐴𝐶 is a secant segment. This means we can use a special
version of the intersecting secants theorem. And that’s called the tangent
secant theorem. In the case of our circle, it tells
us that the product of the lengths of line segments 𝐴𝐵 and 𝐴𝐶 is equal to the
square of the length of line segment 𝐴𝐷.
Now we’re given that 𝐴𝐵 is 12
centimeters, whilst 𝐴𝐶 is 35. So 12 times 35 is equal to 𝐴𝐷
squared, or 𝐴𝐷 squared is equal to 420. We’ll solve this equation by
finding the square root of 420. That gives us that 𝐴𝐷 is 20.493,
which correct the nearest tenth is 20.5 centimeters. The distance from line segment 𝐵𝐶
to the center of the circle 𝑀 is 3.4 centimeters, and the length of line segment
𝐴𝐷 is 20.5 centimeters.
In our next example, we’ll
demonstrate how to use these theorems to set up and solve equations to find missing
values.
In the following figure, find the
value of 𝑥.
And then we have a circle which
contains two chords. Those are chords 𝐴𝐵 and 𝐶𝐷. In fact, those chords intersect at
a point 𝐸 inside the circle. And so we’re going to link the
lengths of each of our respective line segments by using the intersecting chords
theorem. In the case of the given circle,
this tells us that the product of 𝐴𝐸 and 𝐸𝐵 is equal to the product of 𝐶𝐸 and
𝐸𝐷. Well, 𝐴𝐸 times 𝐸𝐵, we see, is
𝑥 plus eight times 𝑥 plus three, whilst 𝐶𝐸 times 𝐸𝐷 can be written as 𝑥 times
𝑥 plus 12.
Let’s distribute all of these
parentheses and then rearrange. The left-hand side of our equation
expands and simplifies to 𝑥 squared plus 11𝑥 plus 24, whilst the right-hand side
simplifies to 𝑥 squared plus 12𝑥. We then observe that we can
subtract 𝑥 squared from both sides of our equation. And we’re left with 11𝑥 plus 24
equals 12𝑥. Next, we’ll subtract 11𝑥 from both
sides. On the left-hand side we’re just
left with 24, whilst on the right we have 12𝑥 minus 11𝑥, which is simply 𝑥. And so, given the information about
our circle, we deduce 𝑥 to be equal to 24.
In our final example, we’ll
demonstrate how to use the intersecting chords theorem in reverse.
Given that 𝐸𝐴 is 5.2 centimeters,
𝐸𝐶 equals six centimeters, 𝐸𝐵 equals 7.5 centimeters, and 𝐸𝐷 equals 6.5
centimeters, do the points 𝐴, 𝐵, 𝐶, and 𝐷 lie on a circle?
And then we’re given a diagram. Now if these points lie on a
circle, then the line segments 𝐴𝐵 and 𝐶𝐷 must themselves be chords to that
circle. And so if these points do indeed
lie on the circumference of the circle, then the respective line segments will
satisfy the intersecting chords theorem. This tells us that the product of
𝐴𝐸 and 𝐸𝐵 must be equal to the product of 𝐷𝐸 and 𝐸𝐶.
So let’s begin by working out the
product of 𝐴𝐸 and 𝐸𝐵. 𝐴𝐸 or 𝐸𝐴 is 5.2 centimeters,
whilst 𝐸𝐵 is 7.5 centimeters. And so their product is 5.2 times
7.5. And that’s equal to 39. Then we want to find the product of
𝐷𝐸 and 𝐸𝐶. Well, 𝐸𝐶 is six centimeters, and
𝐷𝐸, which is equal to 𝐸𝐷, is 6.5 centimeters. So their product is 6.5 times six,
which is also 39. And so we can say, yes, the points
𝐴, 𝐵, 𝐶, and 𝐷 must lie on the circumference of a circle. We might even give a reason and say
that since the line segments satisfy the intersecting chord theorem, 𝐴𝐵 and 𝐶𝐷
must be the chords of the same circle.
We’ll now recap the key points from
this lesson. In this lesson, we learned about
the intersecting chords theorem. And it says that the products of
the lengths of the line segments on each chord must be equal when those two chords
intersect. In other words, in this diagram,
𝐴𝐸 times 𝐸𝐵 equal 𝐶𝐸 times 𝐸𝐷. Similarly, the intersecting secants
theorem says that the product of the measures of one secant segment and its external
secant segment is equal to the product of the measures of the other secant segment
and its external secant segment. In the case of this diagram, 𝐴𝐵
times 𝐵𝐶 equals 𝐴𝐷 times 𝐴𝐸. We also looked at a special case of
this called the tangent secant theorem, which applies where one or both of the lines
are tangent segments.
