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Lesson Explainer: Geometric Constructions: Angle Bisectors Mathematics

In this explainer, we will learn how to construct angle bisectors using rulers and compasses without protractors.

We first note that bisecting is a common term in geometry, and it means to split into two equal parts (or two congruent parts). We can extend this idea to trisecting and beyond. Therefore, bisecting an angle means to split the angle into two congruent angles.

We can define this formally as follows.

Definition: Angle Bisector

The bisector of an angle is the line through the vertex of the angle that splits the angle into two congruent angles.

For example, if we wanted to bisect an angle of 60∘, we would need to split it into two angles, each with a measure of 30∘.

Before we discuss how to bisect an angle in general, let’s see an example of an angle bisector. We recall that a kite 𝐴𝐡𝐢𝐷 is a quadrilateral with two pairs of congruent adjacent sides. We can note that one of the diagonals of any kite splits the shape into two congruent triangles, here △𝐴𝐢𝐷 and △𝐴𝐢𝐡.

Indeed, we can see that both triangles have three congruent sides; hence, the triangles are congruent by the SSS criterion. Therefore, the triangles have congruent angles, so π‘šβˆ π΅π΄πΆ=π‘šβˆ π·π΄πΆ and π‘šβˆ π΅πΆπ΄=π‘šβˆ π·πΆπ΄.

We have shown that the longer diagonal of a kite bisects two of the kite’s internal angles. The same result and proof hold true for rhombuses and actually for both diagonals in a rhombus since they split the rhombus into four congruent triangles.

We can use a similar idea to bisect any angle using a compass and straightedge. Let’s say we have ∠𝐡𝐴𝐢, shown below.

We can trace a circle centered at point 𝐴 that intersects both 𝐴𝐡 and 𝐴𝐢 at points we will label 𝐷 and 𝐸 as shown.

We note that 𝐴𝐸=𝐴𝐷 since these are radii of the same circle. We now want to trace two circles of the same radius centered at 𝐸 and 𝐷 that intersect at a point on the same side as the angle. We can call this point 𝐹 and sketch the lines 𝐸𝐹, 𝐷𝐹, and 𝐴𝐹.

We note that 𝐸𝐹=𝐷𝐹 since they are both radii of circles of the same radius and that 𝐴𝐸=𝐴𝐷 since they are radii of the same circle. Thus, △𝐴𝐹𝐷 and △𝐴𝐹𝐸 have all sides congruent, so △𝐴𝐹𝐷≅△𝐴𝐹𝐸 by the SSS criterion. In particular, this means that their corresponding angles are congruent, and so π‘šβˆ π·π΄πΉ=π‘šβˆ πΈπ΄πΉ. Hence, 𝐴𝐹 bisects ∠𝐡𝐴𝐢.

It is worth noting that this same construction also works for obtuse, reflex, and straight angles. The construction for all three is the same, so we will only show the reflex and straight angle cases.

First, if ∠𝐡𝐴𝐢 were a reflex angle, we once again would trace a circle centered at 𝐴 to find points 𝐷 and 𝐸, which are the points of intersection between the circle and 𝐴𝐢 and 𝐴𝐡, as shown.

We now want to trace two circles centered at 𝐸 and 𝐹 that intersect on the same side as the angle we want to bisect. To do this, we increase the radius of our compass. We call the point of intersection 𝐹 as shown.

We now add in the lines 𝐸𝐹, 𝐷𝐹, and 𝐴𝐹 and note that 𝐴𝐸=𝐴𝐷 and 𝐸𝐹=𝐷𝐹.

Once again, we see that △𝐴𝐹𝐷 and △𝐴𝐹𝐸 have all sides congruent, so △𝐴𝐹𝐷≅△𝐴𝐹𝐸 by the SSS criterion. In particular, this means that their corresponding angles are congruent, and so π‘šβˆ π·π΄πΉ=π‘šβˆ πΈπ΄πΉ. Hence, 𝐴𝐹 bisects the reflex angle ∠𝐡𝐴𝐢.

It is worth noting here that the angle bisector of any angle ∠𝐴𝐡𝐢 and the angle bisector of the reflex angle ∠𝐴𝐡𝐢 are the same straight line. So, we can use either construction.

Similarly, if ∠𝐡𝐴𝐢 is a straight angle, then we trace a circle centered at 𝐴 that intersects 𝐴𝐡 and 𝐴𝐢 to find the points of intersection 𝐷 and 𝐸 as shown.

We then trace two circles of the same radius centered at 𝐷 and 𝐸 that intersect on the same side as the angle we want to bisect. We call this point of intersection 𝐹.

We now add in the lines 𝐸𝐹, 𝐷𝐹, and 𝐴𝐹 and note that 𝐴𝐸=𝐴𝐷 and 𝐷𝐹=𝐸𝐹.

We see that △𝐴𝐹𝐷 and △𝐴𝐹𝐸 have all sides congruent, so △𝐴𝐹𝐷≅△𝐴𝐹𝐸 by the SSS criterion. In particular, this means that their corresponding angles are congruent, and so π‘šβˆ π·π΄πΉ=π‘šβˆ πΈπ΄πΉ. Hence, 𝐴𝐹 bisects the straight angle ∠𝐡𝐴𝐢.

Therefore, we can follow this process to bisect any angle.

How To: Bisecting a Given Angle with a Compass and Straightedge

We can bisect any given angle ∠𝐡𝐴𝐢 with a compass and straightedge using the following construction:

  1. Trace a circle centered at 𝐴 that intersects 𝐴𝐡 and 𝐴𝐢 at distinct points that we will name 𝐷 and 𝐸 respectively.
  2. Trace circles of the same radius centered at 𝐷 and 𝐸 that intersect at a point on the same side as the angle we want to bisect. We will name this point 𝐹.
  3. Sketch line 𝐴𝐹 that bisects ∠𝐡𝐴𝐢.

Let’s now see an example of using this construction to correctly identify what a given geometric construction illustrates.

Example 1: Identifying the Construction Used to Find an Angle Bisector

What does the following figure illustrate?

  1. A bisector of an angle
  2. A perpendicular from a point lying outside a straight line
  3. A bisector of a line segment
  4. A straight line parallel to another line
  5. An angle congruent to another angle

Answer

We start by recalling that tracing a circle at the vertex of an angle and then tracing circles of equal radii centered at those points of intersection such that they intersect at a point on the same side as the angle allows us to bisect the angle by connecting its vertex to the intersection of these circles. Since this is the construction illustrated, the answer is choice A: a bisector of an angle.

We can show this directly from the diagram by labeling the points as follows.

Then, we note that 𝐴𝐢=𝐴𝐡, as they are radii of the same circle, and 𝐢𝐷=𝐡𝐷, as they are radii of congruent circles. Sketching 𝐡𝐷 and 𝐢𝐷 gives us the following.

We note that △𝐴𝐷𝐢 and △𝐴𝐷𝐡 are congruent by the SSS criterion and hence π‘šβˆ π΅π΄π·=π‘šβˆ πΆπ΄π·.

In our next example, we will use this construction of an angle bisector to estimate the length of a section of a line in a triangle.

Example 2: Constructing an Angle Bisector to Find a Length in a Triangle

Given that 𝐴𝐡𝐢 is a triangle, use a ruler and the compass to draw the triangle and bisect ∠𝐢 by the bisector 𝐢𝐷 that intersects 𝐴𝐡 at 𝐷. Use the ruler to measure the length of 𝐴𝐷 to the nearest decimal.

  1. 14.2 cm
  2. 23.7 cm
  3. 33.2 cm
  4. 45.2 cm
  5. 4.7 cm

Answer

We first need to draw triangle 𝐴𝐡𝐢 using a compass and ruler. We will do this by first measuring a straight line of length 5 cm and labeling the endpoints 𝐡 and 𝐢. We then measure the compass to have a radius of 7 cm and trace a circle of radius 7 cm centered at 𝐢. Then, we trace a circle of radius 8 cm centered at 𝐡. The point of intersection between these two circles above 𝐡𝐢 is our point 𝐴. This gives us the following.

We now need to bisect ∠𝐢. We do this by first tracing a circle centered at 𝐢 that intersects both 𝐴𝐢 and 𝐡𝐢. We will call these points 𝐴′ and 𝐡′ as shown.

We now need to trace arcs of two circles of the same radius centered at 𝐴′ and 𝐡′ that intersect at a point on the same side as ∠𝐢. We will label this point 𝐸 as shown.

We now recall that ⃖⃗𝐢𝐸 bisects ∠𝐢. We can extend this line to intersect 𝐴𝐡 and label this point 𝐷 as shown.

We can then measure the length of 𝐴𝐷 using a ruler; we get 4.7 cm to the nearest tenth of a centimetre.

In our next example, we will estimate the perimeter of a triangle constructed using angle bisectors.

Example 3: Constructing an Angle Bisector to Find the Perimeter of a Triangle

Given that 𝐴𝐡𝐢 is a triangle, use a ruler and a compass to draw the triangle shown and bisect ∠𝐢 and ∠𝐡 by the bisectors 𝐢𝐷 and 𝐡𝐷 that intersect at 𝐷. Use the ruler to measure the perimeter of the triangle 𝐷𝐡𝐢 to the nearest decimal.

  1. 19.1 cm
  2. 17.6 cm
  3. 18.4 cm
  4. 16.9 cm
  5. 17.9 cm

Answer

We first need to sketch triangle 𝐴𝐡𝐢 using a ruler and compass. We can do this by first drawing a straight line of length 9 cm and labeling the endpoints 𝐡 and 𝐢. We then set the compass to have a radius of 6 cm and trace a circle of radius 6 cm centered at 𝐢. Next, we set the radius of the compass to be 5 cm and trace a circle of radius 5 cm centered at 𝐡. The point of intersection between these circles above 𝐡𝐢 is point 𝐴.

We now need to find point 𝐷 by bisecting both ∠𝐢 and ∠𝐡 using a compass and straightedge.

Let’s start by bisecting ∠𝐡. We trace a circle centered at 𝐡 that intersects 𝐡𝐴 and 𝐡𝐢 at distinct points 𝐸 and 𝐹 as shown.

We then trace the arcs of two circles of equal radii centered at 𝐸 and 𝐹 that intersect on the same side as ∠𝐡. We label this intersection 𝐺.

We then recall that οƒͺ𝐡𝐺 bisects ∠𝐡.

We follow the same process to bisect ∠𝐢. We trace a circle at 𝐢 to find points on 𝐴𝐢 and 𝐡𝐢, which we label 𝐻 and 𝐼.

We then trace congruent circles centered at 𝐻 and 𝐼 and mark the point of intersection on the same side of ∠𝐢. If this point is labeled 𝐽, we know that 𝐢𝐽 bisects ∠𝐢.

The point of intersection between the bisectors is then point 𝐷.

We need to use a ruler to measure the perimeter of triangle 𝐷𝐡𝐢 to the nearest tenth of a centimetre. We recall that this is the sum of its side lengths, so it is equal to 𝐡𝐢+𝐢𝐷+𝐡𝐷. We note that 𝐡𝐢=9cm. If we measure the sides with a ruler, we get π·π΅β‰ˆ4.2cm and πΆπ·β‰ˆ5.2cm. Adding these together, we get that the perimeter is 18.4 cm to one decimal place.

In the previous example we found the point of intersection between two of the internal angle bisectors of a triangle. Although it is beyond the scope of the explainer to prove the following result, it is worth noting that all of the internal angle bisectors of any triangle intersect at a single point. We can write this formally as follows.

Property: The Angle Bisectors of the Internal Angles of a Triangle Are Concurrent

The internal angle bisectors of any triangle intersect at a single point.

We can see an example of this in the example above if we also draw the angle bisector of ∠𝐴.

Let’s now see an example of using angle bisection and geometric constructions to estimate a length in a trapezoid.

Example 4: Constructing a Right Trapezoid and Using Angle Bisectors to Find a Missing Length

Given that 𝐴𝐡𝐢𝐷 is a trapezoid, use a ruler and a compass to draw the trapezoid and bisect ∠𝐴 by the bisector 𝐴𝑀 that intersects οƒͺ𝐢𝐡 at 𝑀. Use the ruler to measure the length of 𝐴𝑀 to the nearest decimal.

  1. 11.6 cm
  2. 12.0 cm
  3. 13.0 cm
  4. 10.9 cm
  5. 14.1 cm

Answer

We first need to sketch trapezoid 𝐴𝐡𝐢𝐷 using a ruler and compass. We can do this by first drawing a straight line of length 10 cm and labeling the endpoints 𝐴 and 𝐡. We can construct a right angle by bisecting a straight angle at 𝐡. We do this by tracing a circle centered at 𝐡 that intersects ⃖⃗𝐴𝐡 at two points, say 𝐴′ and 𝐡′. Then, we trace circles of equal radii centered at 𝐴′ and 𝐡′ that intersect. The points of intersection of these circles bisect the straight angle into two right angles.

If we measure this line to be 12 cm, then we find point 𝐢.

We now note that 𝐴𝐢𝐷 is a triangle, so we can construct this using a compass and straightedge since we already have points 𝐴 and 𝐢. We set our compass to have a radius of 13 cm and trace a circle of radius 13 cm centered at 𝐴. Then, we set our compass to have a radius of 5 cm and trace a circle of radius 5 cm centered at 𝐢. The point of intersection of these two circles on the correct side of 𝐴𝐢 is then point 𝐷.

We now follow the bisection process to bisect ∠𝐴. We trace a circle centered at 𝐴 and label the points of intersection with 𝐴𝐷 and 𝐴𝐡 𝐸 and 𝐹. Then, we trace circles of equal radii centered at 𝐸 and 𝐹 such that they intersect at a point, 𝐺, on the same side as ∠𝐴 as shown.

We then know that ⃖⃗𝐴𝐺 is the bisector of ∠𝐴. We can sketch this line onto the diagram to find point 𝑀.

We can then measure the length of 𝐴𝑀 using our ruler. To the nearest millimetre, we get 12.0 cm.

In our final example, we will see an example of an interesting geometric property involving angle bisectors in triangles.

Example 5: Observing Symmetries with Angle Bisectors in an Isosceles Triangle

Use a ruler and a compass to draw a triangle 𝐴𝐡𝐢, where 𝐴𝐡=𝐴𝐢=7cm and 𝐡𝐢=6cm, then bisect ∠𝐡 and ∠𝐢 by two bisectors that intersect at 𝑂.

  1. Is 𝑂𝐡 equal to 𝑂𝐢?
  2. Is 𝑂𝐴 equal to 𝑂𝐡?

Answer

Part 1

To construct triangle 𝐴𝐡𝐢, we can start by drawing a line of length 6 cm and labeling the endpoints 𝐡 and 𝐢. We then trace circles of radius 7 cm centered at 𝐡 and 𝐢. We can label either point of intersection between these circles 𝐴.

We now need to bisect ∠𝐡 and ∠𝐢. Let’s start by bisecting ∠𝐡. We trace a circle centered at 𝐡 that intersects 𝐴𝐡 and 𝐢𝐡 at points 𝐷 and 𝐸 as shown.

We then trace circles of equal radii centered at 𝐷 and 𝐸 that intersect on the same side as ∠𝐡. We label this intersection point 𝐹 and recall that ⃖⃗𝐡𝐹 bisects ∠𝐡.

We follow the same process to bisect ∠𝐢. It is worth noting that triangle 𝐴𝐡𝐢 is isosceles, so π‘šβˆ π΅=π‘šβˆ πΆ. This means that the bisected angles will have equal measure. We get the following.

We could then measure the lengths of 𝑂𝐡 and 𝑂𝐢 and see that they are the same. However, we can prove that they are equal by noting that π‘šβˆ πΆπ΅π‘‚=π‘šβˆ π΅πΆπ‘‚, so triangle 𝐡𝐢𝑂 is isosceles.

Hence, we can say that 𝑂𝐡 is equal to 𝑂𝐢.

Part 2

We can follow the same process to bisect ∠𝐴, or we can recall that the bisections of the internal angles of a triangle meet at a point. Either way, we note that ⃖⃗𝑂𝐴 bisects ∠𝐴.

We can measure 𝑂𝐴 to see that it is approximately 4.5 cm, whereas 𝑂𝐡 is approximately 3.5 cm. We can conclude that 𝑂𝐴 is not equal to 𝑂𝐡.

In the previous example, we noted that the distances from the vertex to the intersection point of the bisectors are equal if the angles they bisect are equal. We can use this process to prove that the lengths of the bisectors in an equilateral triangle will all be equal.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The bisector of an angle is the line through the vertex of the angle that splits the angle into two congruent angles.
  • One of the diagonals of a kite bisects two of the kite’s internal angles.
  • We can bisect any given angle ∠𝐡𝐴𝐢 with a compass and straightedge using the following construction:
    1. Trace a circle centered at 𝐴 that intersects 𝐴𝐡 and 𝐴𝐢 at distinct points that we will name 𝐷 and 𝐸 respectively.
    2. Trace circles of the same radius centered at 𝐷 and 𝐸 that intersect at a point on the same side as the angle we want to bisect. We will name this point 𝐹.
    3. Sketch line 𝐴𝐹 that bisects ∠𝐡𝐴𝐢.
  • The angle bisectors of the internal angles in a triangle intersect at a point.

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