# Lesson Explainer: Perpendicular Bisector of a Chord Mathematics

In this explainer, 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 that 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 end points 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. As you can see in the diagram, it also consists of two radii.

In this explainer, we will be considering the perpendicular bisectors of chords, such as the following.

There are three theorems related to perpendicular bisectors of chords, which we will be covering. Let’s start by supposing that we have a chord in a circle, with a straight line that passes through the center of the circle, , and also bisects the chord, , as shown in the diagram.

We will now prove by contradiction that the bisector of the chord, which passes through the center of the circle, is also perpendicular to the chord. We will assume that and .

In our diagram, we can see that we have a circle centered at , the chord , the bisector , and the two radii and .

Now, since and are radii of the circle, they will be equal in length. Also, since bisects , we can say that the lengths of and will be equal. Finally, and share the side , so we can see that all three side of these triangles are equal in length. Therefore, by SSS congruence, we have that

Next, we consider the angles and . We know that is a straight line because it is a chord within the circle. Angles on a straight line sum to . Due to the congruence of and , we have that . Combining these two points, we have that

This contradicts our assumption at the beginning and, hence, completes our proof. Let’s summarize what we have just proven with the following theorem.

### Theorem: The Chord Bisector Theorem—Part 1

If we have a circle with center containing a chord , then the straight line that passes through and bisects chord is perpendicular to .

The second chord bisector theorem is very similar to this one. However, we start with a slightly different assumption. This time, we have a circle containing a chord, with a straight line passing through the center of the circle, , which is also perpendicular to chord as shown in this diagram.

The theorem states that the line that passes through the center of the circle and is perpendicular to the chord also bisects that chord. The proof of this theorem is very similar to the first proof, and so we will not be proving it here.

### Theorem: The Chord Bisector Theorem—Part 2

If we have a circle with center containing a chord , then the straight line that passes through and is perpendicular to also bisects .

The final theorem we will be covering is often called the converse to the chord bisector theorem. It is very similar to the other two theorems but again starts with slightly different assumptions. This theorem states that the perpendicular bisector of a chord also passes through the center of the circle. The proof is also very similar to the proof of the first theorem, and so this is left as an exercise for the reader.

### Theorem: The Converse to the Chord Bisector Theorem

If we have a circle with center containing a chord , then the perpendicular bisector of passes through .

Now that we have covered these three theorems, let us look at some examples of how we can use them to find missing lengths and angles.

### Example 1: Finding a Missing Length Using Perpendicular Bisectors of Chords

Given and , find .

### Answer

In the diagram, we can see that we have a circle with center . There is a chord , which is bisected by the line segment at point . Here, we can apply the chord bisector theorem, which states that if we have a circle with center containing a chord , then the straight line that passes through and bisects chord is perpendicular to . Hence, we can say that

The question tells us that . is a radius of the circle, and so is ; therefore,

We have also been given that , so let’s add these to our diagram.

We can see in our diagram that is a right triangle of which we know two of the sides. We can find the remaining side using the Pythagorean theorem:

All that is left to do now is to use the fact that bisects , so . Substituting our value for in, we obtain that

In the next example, we will see how we can apply the theorems to find the area of a triangle.

### Example 2: Finding a Missing Length and the Area of a Triangle Using Perpendicular Bisectors of Chords

In the figure below, if and , find the length of and the area of to the nearest tenth.

### Answer

Since is the center of the circle and line bisects chord at , we can apply the chord bisector theorem, which states that if we have a circle with center containing a chord , then the straight line that passes through and bisects chord is perpendicular to . Hence, we can say that

We have been given the length of , and we also know that bisects this chord, so we can say that

Adding this information to our diagram, we can see that we have a right triangle, , for which we know two of the side lengths. We can use the Pythagorean theorem to find the length of :

To answer the first part of the question, we need to round this value to the nearest tenth. In doing this, we obtain

Next, we need to find the area of . We know that is perpendicular to , so we have that . We also have that , so all we need to find is the length of .

and are both radii of the circle, so they will have the same length. Therefore, . We can find the length of by subtracting the length we found previously for from the length of . In doing this, we obtain

We know that the area of a triangle is given by the formula

Using this formula, we find that the area of this triangle is

All we need to do now is to round to the nearest tenth, to find that our solution is

In our next example, we will see how we can use perpendicular bisectors of chords to find a missing angle.

### Example 3: Finding a Missing Angle Using Perpendicular Bisectors of Chords

and are two chords in a circle with center on two opposite sides of the center, where . If and are the midpoints of and , respectively, find .

### Answer

We can start our solution by noting that both and pass through the center of the circle and bisect chords and respectively. Therefore, we can apply the chord bisector theorem, which states that if we have a circle with center containing a chord , then the straight line that passes through and bisects chord is perpendicular to . Using this theorem, we can say that and , so we have that

Now, if we consider quadrilateral , we can see that we now know the measures of three of the interior angles. Since the interior angles of a quadrilateral always sum to , we can find the measure of the unknown angle as follows:

In the next example, we will find the perimeter of a triangle using perpendicular bisectors of chords.

### Example 4: Finding the Perimeter of a Triangle Drawn inside Another Triangle Whose Vertices Touch a Circle

In a circle of center , , , and . Given that and , find the perimeter of .

### Answer

The first thing we notice about this question is that line segments and both pass through and meet chords and , respectively, at right angles. Therefore, we can apply the theorem, which states that if we have a circle with center containing a chord , then the straight line that passes through and is perpendicular to also bisects .

Using this information, we can say that

This tells us that is the midpoint of and is the midpoint of .

Now we will use the midpoint theorem, which tells us that the line segment in a triangle joining the midpoints of two sides of the triangle is said to be parallel to the third side and is also half the length of the third side. Hence, we have that

We also know that and . By summing these three lengths, we will reach our solution:

In our final example, we will see how we can use the perpendicular bisectors to find the diameter of a circle, given the length of a chord.

### Example 5: Determining the Length of the Diameter of a Circle given the Length of a Chord in It

Using the figure and the fact that , determine the diameter of the circle.

### Answer

The question has asked us to find the length of the diameter of the circle. Since the diameter is double of the radius, a sensible place to start will be to identify the circle’s radius.

The first thing we notice when looking at this question is that the line segment passes through the center of the circle and is also perpendicular to chord . This tells us that bisects . Since , we can say that

Next, let’s consider . We know that , but we also know that , since they are both radii of the circle. Hence, this triangle is an equilateral triangle. Using this, we can say that

The following diagram shows the information we have found so far.

We will now consider . We can see that it is a right triangle, for which we know and . There are two methods we can use to find the length of , which is also the radius of the circle. We will cover both methods.

Method 1

We will start by considering . We know that the angles in a triangle sum to and the other two angles in are and . Therefore, we have that

Now we can use the fact that in a 30-60-90 triangle, the side that is opposite the angle of is half the length of the hypotenuse, so .

Since this is a right triangle, we can use the Pythagorean theorem, which tells us that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Applying this to our triangle gives us

Next, we can substitute in and and then simplify:

We note that when we take the square root in the final step here, we can ignore the negative solution since the length, , must be positive. Now we can find the length of using . This give us that

This is the radius of the circle. However, the question asked us to find the diameter, so we need to double this to reach our solution of

Method 2

In the second method, we will be using right triangle trigonometry. We know that

In our case, the opposite side is and the hypotenuse is . We can rearrange this formula to obtain

Now that we have found the length of the radius, all that is left to do is to double it to find the length of the diameter. This gives us a solution of

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. Let’s recap some key points of the explainer.

### Key Points

• If we have a circle with center containing a chord , then the straight line that passes through and bisects chord is perpendicular to .
• If we have a circle with center containing a chord , then the straight line that passes through and is perpendicular to also bisects .
• If we have a circle with center containing a chord , then the perpendicular bisector of passes through .
• These theorems can be used to find missing lengths, angle measures, and other unknowns in problems involving circles.

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