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, 𝐴, and also bisects the chord, 𝐡𝐢, 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 βˆ π΄π·π΅β‰ 90∘ and βˆ π΄π·πΆβ‰ 90∘.

In our diagram, we can see that we have a circle centered at 𝐴, the chord 𝐡𝐢, the bisector ⃖⃗𝐸𝐹, and the two radii 𝐴𝐡 and 𝐴𝐢.

Now, since 𝐴𝐡 and 𝐴𝐢 are radii of the circle, they will be equal in length. Also, since ⃖⃗𝐸𝐹 bisects 𝐢𝐡, we can say that the lengths of 𝐢𝐷 and 𝐷𝐡 will be equal. Finally, △𝐴𝐢𝐷 and △𝐴𝐷𝐡 share the side 𝐴𝐷, so we can see that all three side of these triangles are equal in length. Therefore, by SSS congruence, we have that △𝐴𝐢𝐷≅△𝐴𝐷𝐡.

Next, we consider the angles ∠𝐴𝐷𝐢 and ∠𝐴𝐷𝐡. We know that 𝐢𝐡 is a straight line because it is a chord within the circle. Angles on a straight line sum to 180∘. Due to the congruence of △𝐴𝐢𝐷 and △𝐴𝐷𝐡, we have that ∠𝐴𝐷𝐡=∠𝐴𝐷𝐢. Combining these two points, we have that ∠𝐴𝐷𝐡=∠𝐴𝐷𝐢=1802=90.∘∘

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 Theoremβ€”Part 1

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

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, 𝐴, which is also perpendicular to chord 𝐡𝐢 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 Theoremβ€”Part 2

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

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 𝐴 containing a chord 𝐡𝐢, then the perpendicular bisector of 𝐡𝐢 passes through 𝐴.

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 𝐴𝑀=200cm and 𝑀𝐢=120cm, find 𝐴𝐡.

Answer

In the diagram, we can see that we have a circle with center 𝑀. There is a chord 𝐴𝐡, which is bisected by the line segment 𝑀𝐷 at point 𝐢. Here, we can apply the chord bisector theorem, which states that if we have a circle with center 𝑀 containing a chord 𝐴𝐡, then the straight line that passes through 𝑀 and bisects chord 𝐴𝐡 is perpendicular to 𝐴𝐡. Hence, we can say that π‘šβˆ π‘€πΆπ΅=90.∘

The question tells us that 𝐴𝑀=200cm. 𝐴𝑀 is a radius of the circle, and so is 𝐡𝑀; therefore, 𝐡𝑀=200.cm

We have also been given that 𝑀𝐢=120cm, so let’s add these to our diagram.

We can see in our diagram that △𝑀𝐢𝐡 is a right triangle of which we know two of the sides. We can find the remaining side using the Pythagorean theorem: 𝐡𝐢=√200βˆ’120=160.cm

All that is left to do now is to use the fact that 𝐢 bisects 𝐴𝐡, so 𝐴𝐡=2𝐡𝐢. Substituting our value for 𝐡𝐢 in, we obtain that 𝐴𝐡=320.cm

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 𝑀𝐴=17.2cm and 𝐴𝐡=27.6cm, find the length of 𝑀𝐢 and the area of △𝐴𝐷𝐡 to the nearest tenth.

Answer

Since 𝑀 is the center of the circle and line 𝑀𝐷 bisects chord 𝐴𝐡 at 𝐢, we can apply the chord bisector theorem, which states that if we have a circle with center 𝑀 containing a chord 𝐴𝐡, then the straight line that passes through 𝑀 and bisects chord 𝐴𝐡 is perpendicular to 𝐴𝐡. Hence, we can say that π‘šβˆ π‘€πΆπ΄=90.∘

We have been given the length of 𝐴𝐡, and we also know that 𝐢 bisects this chord, so we can say that 𝐴𝐢=27.62=13.8.cm

Adding this information to our diagram, we can see that we have a right triangle, △𝑀𝐢𝐴, for which we know two of the side lengths. We can use the Pythagorean theorem to find the length of 𝑀𝐢: 𝑀𝐢=√17.2βˆ’13.8=10.266….cm

To answer the first part of the question, we need to round this value to the nearest tenth. In doing this, we obtain 𝑀𝐢=10.3.cm

Next, we need to find the area of △𝐴𝐷𝐡. We know that 𝑀𝐷 is perpendicular to 𝐴𝐡, so we have that π‘šβˆ π΄πΆπ·=90∘. We also have that 𝐴𝐡=27.6cm, so all we need to find is the length of 𝐢𝐷.

𝑀𝐷 and 𝑀𝐴 are both radii of the circle, so they will have the same length. Therefore, 𝑀𝐷=17.2cm. We can find the length of 𝐢𝐷 by subtracting the length we found previously for 𝑀𝐢 from the length of 𝑀𝐷. In doing this, we obtain 𝐢𝐷=17.2βˆ’10.266…=6.933….cm

We know that the area of a triangle is given by the formula areaofatrianglebaseheight=12Γ—Γ—.

Using this formula, we find that the area of this triangle is areaofcm△𝐴𝐷𝐡=12Γ—27.6Γ—6.933…=95.683….

All we need to do now is to round to the nearest tenth, to find that our solution is areaofcm△𝐴𝐷𝐡=95.7.

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

𝐴𝐡 and 𝐴𝐢 are two chords in a circle with center 𝑀 on two opposite sides of the center, where π‘šβˆ π΅π΄πΆ=33∘. If 𝐷 and 𝐸 are the midpoints of 𝐴𝐡 and 𝐴𝐢, respectively, find π‘šβˆ π·π‘€πΈ.

Answer

We can start our solution by noting that both 𝑀𝐸 and 𝑀𝐷 pass through the center of the circle and bisect chords 𝐴𝐢 and 𝐴𝐡 respectively. Therefore, we can apply the chord bisector theorem, which states that if we have a circle with center 𝑀 containing a chord 𝐴𝐡, then the straight line that passes through 𝑀 and bisects chord 𝐴𝐡 is perpendicular to 𝐴𝐡. Using this theorem, we can say that π‘€π·βŸ‚π΄π΅ and π‘€πΈβŸ‚π΄πΆ, so we have that π‘šβˆ π‘€πΈπ΄=90π‘šβˆ π‘€π·π΄=90.∘∘and

Now, if we consider quadrilateral 𝐴𝐷𝑀𝐸, 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 360∘, we can find the measure of the unknown angle as follows: π‘šβˆ π·π‘€πΈ=360βˆ’90βˆ’90βˆ’33=147.∘

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 𝑂, 𝐴𝐡=35cm, 𝐢𝐡=25cm, and 𝐴𝐢=40cm. Given that π‘‚π·βŸ‚π΅πΆ and π‘‚πΈβŸ‚π΄πΆ, find the perimeter of △𝐢𝐷𝐸.

Answer

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

Using this information, we can say that 𝐴𝐸=𝐢𝐸𝐢𝐷=𝐷𝐡.and

This tells us that 𝐸 is the midpoint of 𝐴𝐢 and 𝐷 is the midpoint of 𝐡𝐢.

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 𝐸𝐷=12𝐴𝐡=352=17.5.cm

We also know that 𝐢𝐸=12𝐴𝐢=20cm and 𝐢𝐷=12𝐢𝐡=12.5cm. By summing these three lengths, we will reach our solution: theperimeterofcm△𝐢𝐷𝐸=17.5+20+12.5=50.

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 𝐡𝐢=32√3cm, 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 𝑀𝐴 passes through the center of the circle and is also perpendicular to chord 𝐡𝐢. This tells us that 𝑀𝐴 bisects 𝐡𝐢. Since 𝐡𝐢=32√3cm, we can say that 𝐡𝐷=16√3.cm

Next, let’s consider △𝑀𝐡𝐴. We know that 𝑀𝐡=𝐡𝐴, but we also know that 𝑀𝐡=𝑀𝐴, since they are both radii of the circle. Hence, this triangle is an equilateral triangle. Using this, we can say that π‘šβˆ π΄π‘€π΅=60.∘

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

We will now consider △𝑀𝐷𝐡. We can see that it is a right triangle, for which we know βˆ π·π‘€π΅=60∘ and 𝐡𝐷=16√3cm. There are two methods we can use to find the length of 𝑀𝐡, which is also the radius of the circle. We will cover both methods.

Method 1

We will start by considering βˆ π‘€π΅π·. We know that the angles in a triangle sum to 180∘ and the other two angles in △𝑀𝐷𝐡 are 90∘ and 60∘. Therefore, we have that βˆ π‘€π΅π·=180βˆ’90βˆ’60=30.∘

Now we can use the fact that in a 30-60-90 triangle, the side that is opposite the angle of 30∘ is half the length of the hypotenuse, so 𝑀𝐡=2𝑀𝐷.

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 𝑀𝐡=𝑀𝐷+𝐡𝐷.

Next, we can substitute in 𝑀𝐡=2𝑀𝐷 and 𝐡𝐷=16√3cm and then simplify: (2𝑀𝐷)=𝑀𝐷+ο€»16√33𝑀𝐷=768𝑀𝐷=256𝑀𝐷=16.cm

We note that when we take the square root in the final step here, we can ignore the negative solution since the length, 𝑀𝐷, must be positive. Now we can find the length of 𝑀𝐡 using 𝑀𝐡=2𝑀𝐷. This give us that 𝑀𝐡=32.cm

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 64.cm

Method 2

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

In our case, the opposite side is 𝐡𝐷 and the hypotenuse is 𝑀𝐡. We can rearrange this formula to obtain 𝑀𝐡=16√360=32.sincm∘

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 64.cm

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

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