Video Transcript
In this video, we will learn how to
use the theory of the perpendicular bisector of a chord from the center of a circle
and its converse to solve problems. Let’s begin by recalling how we can
define a radius, a chord, and a diameter. A radius is a line segment which
has one end at the center of the circle and the other on the circumference. We define a chord as any straight
line segment whose endpoints both lie on the circumference of the same circle. The diameter is a special type of
chord which passes through the center of the circle. We can also consider this to be
made up of two radii.
We will now consider what a
perpendicular bisector of a chord looks like. In the diagram drawn, we have the
chord 𝐵𝐶 together with its perpendicular bisector. We will look at three theorems in
this video. And in each case, we need to
consider the center of the circle 𝐴 together with the radii 𝐴𝐵 and 𝐴𝐶.
Our first theorem states that if we
have a circle with center 𝐴 containing a chord 𝐵𝐶, then the straight line that
passes through 𝐴 and bisects the chord 𝐵𝐶 is perpendicular to 𝐵𝐶. The second theorem is very
similar. If we have a circle with center 𝐴
containing a chord 𝐵𝐶, then the straight line that passes through 𝐴 and is
perpendicular to 𝐵𝐶 also bisects 𝐵𝐶. Our third theorem is the converse
to the chord bisector theorem. This states that if we have a
circle with center 𝐴 containing a chord 𝐵𝐶, then the perpendicular bisector of
𝐵𝐶 passes through 𝐴.
It is important to note that the
perpendicular bisector of a chord creates two congruent triangles. In the diagram drawn, these are
triangles 𝐴𝐷𝐶 and 𝐴𝐷𝐵. We will now look at some examples
where we can use the theorems discussed to find missing lengths and angles.
Given 𝐴𝑀 equals 200 centimeters
and 𝑀𝐶 equals 120 centimeters, find the length of the line segment 𝐴𝐵.
In the diagram shown, we have a
circle with center 𝑀. The chord 𝐴𝐵 is bisected by the
line segment 𝑀𝐷 at the point 𝐶. Applying the chord bisector
theorem, which states that if we have a circle with center 𝑀 containing a chord
𝐴𝐵, then the straight line that passes through 𝑀 and bisects the chord 𝐴𝐵 is
perpendicular to 𝐴𝐵. We can say that the measure of
angle 𝑀𝐶𝐵 is 90 degrees. We are told that the length of 𝐴𝑀
is 200 centimeters. And since this is a radius of the
circle, 𝐵𝑀 is also 200 centimeters. We are also told that 𝑀𝐶 is equal
to 120 centimeters. Adding these measurements to our
diagram, we have a right triangle 𝑀𝐶𝐵 as shown.
Applying the Pythagorean theorem,
the length of 𝐵𝐶 is equal to the square root of 200 squared minus 120 squared. This is equal to 160. As 𝐵𝐶 is equal to 160 centimeters
and the point 𝐶 bisects the line segment 𝐴𝐵, then 𝐴𝐶 is also equal to 160
centimeters. The line segment 𝐴𝐵 is therefore
equal to 320 centimeters.
In our next example, we will see
how we can apply the theroems to find the area of a triangle.
In the figure below, if 𝑀𝐴 is
equal to 17.2 centimeters and 𝐴𝐵 is equal to 27.6 centimeters, find the length of
the line segment 𝑀𝐶 and the area of triangle 𝐴𝐷𝐵 to the nearest tenth.
Since 𝑀 is the center of the
circle and the line segment 𝑀𝐷 bisects the chord 𝐴𝐵 at 𝐶, we can apply the
chord bisector theorem. This states that if we have a
circle with center 𝑀 containing a chord 𝐴𝐵, then the straight line that passes
through 𝑀 and bisects 𝐴𝐵 is perpendicular to 𝐴𝐵. This means that the measure of
angle 𝑀𝐶𝐴 is 90 degrees. Since the chord 𝐴𝐵 has length
27.6 centimeters and we know that 𝐶 bisects this chord, 𝐴𝐶 is equal to 27.6
divided by two. This is equal to 13.8
centimeters.
We are also told that the radius
𝑀𝐴 is equal to 17.2 centimeters. We can therefore use the
Pythagorean theorem in the right triangle 𝑀𝐶𝐴 such that 𝑀𝐶 is equal to the
square root of 17.2 squared minus 13.8 squared. This is equal to 10.266 and so
on. And rounding to the nearest tenth,
the line segment 𝑀𝐶 is equal to 10.3 centimeters.
The second part of our question
asks us to calculate the area of triangle 𝐴𝐷𝐵. Clearing some space, we recall that
the area of any triangle is equal to the length of its base multiplied by the length
of its perpendicular height divided by two. We know that the base of our
triangle 𝐴𝐵 is equal to 27.6 centimeters. The perpendicular height 𝐶𝐷 will
be equal to the length of 𝑀𝐷 minus the length of 𝑀𝐶. 𝑀𝐷 is the radius of the circle,
and we know this is equal to 17.2 centimeters. Whilst we could use 10.3
centimeters for 𝑀𝐶, it is better for accuracy to use the nonrounded version: 𝑀𝐶
is equal to 10.266 and so on. Subtracting this from 17.2, we see
that the length of 𝐶𝐷 is 6.933 and so on centimeters.
We can now calculate the area of
triangle 𝐴𝐷𝐵 by multiplying this by 27.6 centimeters and then dividing by
two. This is equal to 95.683 and so
on. Once again, we need to round to the
nearest tenth, giving us an answer of 95.7 square centimeters.
In our next example, we will see
how we can use perpendicular bisectors of chords to find a missing angle.
Line segments 𝐴𝐵 and 𝐴𝐶 are two
chords in the circle with center 𝑀 in two opposite sides of its center, where the
measure of angle 𝐵𝐴𝐶 is 33 degrees. If 𝐷 and 𝐸 are the midpoints of
the line segments 𝐴𝐵 and 𝐴𝐶, respectively, find the measure of angle 𝐷𝑀𝐸.
We begin by noticing that 𝑀𝐸 and
𝑀𝐷 both pass through the center of the circle and that they bisect the chords 𝐴𝐶
and 𝐴𝐵, respectively. We can therefore apply the chord
bisector theorem, which states if we have a circle with center 𝑀 containing a chord
𝐴𝐵, then the straight line which passes through 𝑀 and bisects 𝐴𝐵 is
perpendicular to 𝐴𝐵. This means that, on our diagram,
the measure of angle 𝑀𝐸𝐴 and the measure of angle 𝑀𝐷𝐴 are both equal to 90
degrees.
We notice that 𝐴𝐷𝑀𝐸 is a
quadrilateral. And we know that the angles in a
quadrilateral sum to 360 degrees. This means that the measure of
angle 𝐷𝑀𝐸 which we are trying to calculate is equal to 360 minus 90 minus 90
minus 33. This is equal to 147 degrees.
In our final example, we will find
the perimeter of a triangle using perpendicular bisectors of chords.
In a circle of center 𝑂, 𝐴𝐵 is
equal to 35 centimeters, 𝐶𝐵 is equal to 25 centimeters, and 𝐴𝐶 is equal to 40
centimeters. Given that line segment 𝑂𝐷 is
perpendicular to line segment 𝐵𝐶 and line segment 𝑂𝐸 is perpendicular to line
segment 𝐴𝐶, find the perimeter of triangle 𝐶𝐷𝐸.
We are given in the question the
length of the three sides of the triangle 𝐶𝐵𝐴. We know that 𝐴𝐵 is 35
centimeters, 𝐶𝐵 is 25 centimeters, and 𝐴𝐶 is 40 centimeters. We have been asked to calculate the
perimeter of triangle 𝐶𝐷𝐸. We will do this by firstly proving
that triangles 𝐶𝐵𝐴 and 𝐶𝐷𝐸 are similar using the chord bisector theorem. We notice from the diagram that the
line segments 𝑂𝐸 and 𝑂𝐷 both pass through 𝑂 and meet the chords 𝐴𝐶 and 𝐶𝐵
at right angles.
The chord bisector theorem states
that if we have a circle with center 𝑂 containing a chord 𝐵𝐶, then the straight
line that passes through 𝑂 and is perpendicular to 𝐵𝐶 also bisects 𝐵𝐶. In our diagram, this means that the
length of 𝐴𝐸 is equal to the length 𝐸𝐶 and the length 𝐶𝐷 is equal to the
length 𝐷𝐵.
It is also clear from the diagram
that 𝐴𝐶 is equal to two multiplied by 𝐸𝐶 and 𝐶𝐵 is equal to two multiplied by
𝐶𝐷. As the two triangles 𝐶𝐵𝐴 and
𝐶𝐷𝐸 also share the angle 𝐶, we have two corresponding sides in proportion and
the angle between the two sides is congruent. This proves that the two triangles
are similar. And in fact triangle 𝐶𝐵𝐴 is
larger than triangle 𝐶𝐷𝐸 by a scale factor of two, as the lengths of the
corresponding sides are twice as long. Side 𝐴𝐶 is equal to two
multiplied by side 𝐸𝐶, 𝐶𝐵 is equal to two 𝐶𝐷, and 𝐴𝐵 is equal to two
multiplied by 𝐸𝐷.
We can calculate the perimeter of
triangle 𝐶𝐵𝐴 by adding 40, 35, and 25. This is equal to 100
centimeters. The perimeter of triangle 𝐶𝐷𝐸
will therefore be equal to half of this. This is equal to 50
centimeters.
We have now seen a variety of
examples of how perpendicular bisectors of chords can be used to find missing
lengths, angle measures, and other unknowns in problems involving circles. We will now recap the key points
from this video.
The chord bisector theorem can be
summarized in three ways. Firstly, if we have a circle with
center 𝐴 containing a chord 𝐵𝐶, then the straight line that passes through 𝐴 and
bisects the chord 𝐵𝐶 is perpendicular to 𝐵𝐶. In the same way, the straight line
that passes through 𝐴 and is perpendicular to 𝐵𝐶 also bisects 𝐵𝐶. The converse of these two states
that the perpendicular bisector of the chord 𝐵𝐶 passes through the center 𝐴. As already stated, these theorems
can be used to find missing lengths, angle measures, and other unknowns in problems
involving circles.
