Lesson Video: Geometric Constructions: Angle Bisectors Mathematics • First Year of Preparatory School

Video Transcript

Geometric Constructions: Angle Bisectors

In this lesson, we will learn about the bisector of an angle and how to construct the angle bisector of any given angle using a pair of compasses and a straightedge.

We can start by describing what we mean by the bisector of an angle. An angle bisector is the line through the vertex of an angle splitting it into two congruent angles. We can say that the angle is bisected since it is split in half. We can extend this idea to angle trisectors, which are lines that split an angle into thirds, and we can extend this idea even further. However, we will only focus on bisectors in this video.

We can see many examples of angle bisectors by considering angle measures. For instance, let’s say that we have an angle of measure 60 degrees, as shown. We know that the angle bisector is the line through the vertex of the angle that splits the angle of measure 60 degrees into two congruent angles. Since half of 60 is 30, we know that the measure of the two angles after splitting in half must be 30 degrees. This gives us the following line that is the angle bisector. It is worth noting that we usually only draw the line as a ray starting at the vertex on the same side as the angle. However, we can draw this as a line as well.

It is worth reiterating at this point that we do not want to rely on angle measure to find angle bisectors. For example, using a protractor to split an angle in half will have measurement errors. Instead, we want to find a method to accurately bisect any given angle. To do this, we first need to recall a property about kites, which are quadrilaterals with pairs of congruent adjacent sides as shown. If we construct the diagonal between the pairs of congruent sides, in this case it’s the line 𝐴𝐶, then we form two triangles: 𝐴𝐵𝐶 and 𝐴𝐷𝐶. We can then note that these two triangles have the same side lengths. So they are congruent by the side-side-side congruency criterion.

By using this congruence, we can conclude that the corresponding angles are also congruent. In particular, we can see that angle 𝐶𝐴𝐵 and angle 𝐶𝐴𝐷 have the same measure. Hence, the diagonal is the bisector of this angle.

It is worth noting that this result and proof also works for a rhombus, that is, a quadrilateral with all sides the same length. In fact, both diagonals of a rhombus bisect its internal angles.

We can use this idea to bisect any angle with a pair of compasses and straightedge. We need to construct a kite or rhombus at the vertex of the angle using the sides of the angle and then use the diagonal of the shape to bisect the angle.

To see how we can do this, let’s bisect the given acute angle 𝐶𝐴𝐵. We can construct a rhombus using the sides of the angle by using properties of circles. First, let’s trace a circle centered at 𝐴 that intersects the sides of the angle at two points 𝐷 and 𝐸 as shown. Since line segments 𝐴𝐸 and 𝐴𝐷 are radii of the same circle, we can note that they are the same length. We want to form a quadrilateral with all four sides the same length. So we need to find another point of this same distance from 𝐷 and 𝐸. We can do this by tracing congruent circles at 𝐷 and 𝐸 and labeling the points of intersection between the circles 𝐹 as shown.

Now, we can note that quadrilateral 𝐴𝐸𝐹𝐷 is a rhombus, since all of its sides are the same length. We know that they are the same length since they are all radii of congruent circles.

Finally, we recall that the diagonal of a rhombus bisects its interior angles. So we can conclude that the line between 𝐴 and 𝐹 must be the bisector of angle 𝐶𝐴𝐵.

Before we consider this construction for angles that are not acute, let’s write down the steps we took for constructing the bisector of angle 𝐵𝐴𝐶 using a pair of compasses and a straightedge. First, we trace a circle centered at the vertex of the angle 𝐴 that intersects the sides of the angle 𝐴𝐵 and 𝐴𝐶 at two distinct points we label 𝐷 and 𝐸, respectively. Second, we want to trace congruent circles centered at points 𝐷 and 𝐸. We then label the point of intersection between these two circles that lies on the same side as the angle we want to bisect 𝐹. Finally, we can sketch the line, ray, or line segment between 𝐴 and 𝐹 and note that it is the diagonal of a rhombus, with the sides of the angle as its sides, so it must bisect the angle.

This construction and the proof of the construction work in more or less the same way for obtuse, reflex, and straight angles. To see this, let’s go through an example of using this process to bisect the reflex angle 𝐶𝐴𝐵 shown. We start by tracing a circle centered at the vertex of the angle 𝐴 that intersects the sides of the angle at two distinct points 𝐷 and 𝐸 as shown. Second, we need to trace congruent circles centered at 𝐷 and 𝐸 that intersect at a point on the same side as the angle. To do this, we need to increase the radius of our pair of compasses. Doing this, we obtain a point like 𝐹 as shown.

It is worth noting that since we increased the radius of our compass, we no longer have a rhombus. We can however note that quadrilateral 𝐹𝐸𝐴𝐷 is a kite, so the proof remains the same.

Finally, we sketch the line between 𝐴 and 𝐹 and note that it is the diagonal of a kite between its pair of congruent sides. So it is the bisector of the reflex angle 𝐵𝐴𝐶. Another way of seeing why this holds true is to draw in the lines between 𝐸 and 𝐹 and 𝐷 and 𝐹. We can then see that triangle 𝐴𝐹𝐸 is congruent to triangle 𝐴𝐹𝐷 by the side-side-side congruency criterion. Let’s now see an example of applying this process to identify a given geometric construction.

What does the following figure illustrate? Option (A) a bisector of an angle. Option (B) a perpendicular from a point lying outside a straight line. Option (C) a bisector of a line segment. Option (D) a straight line parallel to another line. Or is it option (E) an angle congruent to another angle?

In this question, we are given five options and we need to determine which of the options is illustrated by the figure. We can directly answer this question by recalling the construction of an angle bisector. However, we can also do this by using the figure. We first note that a circle is traced at the vertex of the angle. So the radii of this circle must have the same length. We can then note that congruent circles are traced centered at the points of intersection of the sides of the angle and the original circle. This means that we can note that radii of these circles have the same length. This gives us the following kite.

Finally, either by using the fact that the diagonal of a kite between its pair of congruent sides bisects its interior angles or by using the congruency of two triangles, we can conclude that the diagram shows the bisector of an angle.

In our next example, we will construct the bisector of an interior angle in a triangle in order to find a point on the side of the triangle so that we can compare the ratio of the lengths the side is split into.

Given that 𝐴𝐵𝐶 is a triangle, use a ruler and a pair of compasses to draw the triangle and bisect angle 𝐶 by the bisector the ray from 𝐶 through 𝐷 that intersects the ray from 𝐴 through 𝐵 at 𝐷. Use the ruler to measure the length of the line segment 𝐴𝐷 to the nearest one decimal place. Option (A) 4.2 centimeters, option (B) 3.2 centimeters, option (C) 4.7 centimeters, option (D) 3.7 centimeters. Or is it option (E) 5.2 centimeters?

In this question, we are asked to draw a triangle of given lengths using a ruler and a pair of compasses, bisect the angle at 𝐶 in a way to find a point 𝐷, and then measure the length of the line segment 𝐴𝐷 to the nearest tenth of a centimeter.

Let’s start with the first part of the question. We are asked to draw the triangle with the given lengths. We might be tempted to skip this step and use the given diagram. However, these are not always accurate. So it is a good idea to construct the triangle ourselves, although in this case the diagram is accurate.

To construct a triangle with a compass and ruler, we need to start by constructing one side of the triangle. Let’s say we start with 𝐵𝐶, which is of length five centimeters. We can then trace a circle of radius seven centimeters centered at 𝐶 and a circle of radius eight centimeters centered at 𝐵 and label either point of intersection between the circles 𝐴. We can then use the radius of each circle to note that triangle 𝐴𝐵𝐶 has the desired side lengths.

The second part of this question asks us to bisect the angle at 𝐶, that is, the internal angle of the triangle. We can recall that the bisector will split this angle into two congruent angles. We can also note that this bisector will lie inside the triangle, so point 𝐷 will lie on the line segment 𝐴𝐵. We do not need to extend this side. We then recall that we can bisect angles by a construction. First, we trace a circle centered at the vertex of the angle we want to bisect and label the points of intersection between the circle and the sides 𝐴 prime, 𝐵 prime as shown. We can then sketch congruent circles centered at 𝐴 prime and 𝐵 prime that intersect at a point on the same side as the angle we want to bisect. We will call this point of intersection 𝐸.

We can then conclude that the line between 𝐶 and 𝐸 bisects the angle at 𝐶. We can extend this bisector to intersect the opposite side and label the point of intersection 𝐷 as shown. We can then measure the length of the line segment 𝐴𝐷 with our ruler. And if we do, we get 4.7 centimeters to the nearest tenth of a centimeter, which is option (C).

It is worth noting that this shows that the angle bisector in a triangle does not necessarily bisect the opposite side. Another useful application of bisecting angles is to consider the bisection of a straight angle, say 𝐵𝐴𝐶. We trace a circle centered at 𝐴 and label the points of intersection 𝐷 and 𝐸 and then trace congruent circles centered at 𝐷 and 𝐸 and label the point of intersection 𝐹 as shown. We then know that 𝐴𝐹 is the bisector of the straight angle, so it must split the angle into two right angles. This means that we can use this construction to construct right angles with a pair of compasses and a straightedge. We can even apply this process a second time to bisect the right angle, giving us a process of constructing a 45-degree angle with a compass and straightedge.

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

Given that 𝐴𝐵𝐶 is a triangle, use a ruler and a pair of compasses to draw the triangle shown and bisect angle 𝐶 and angle 𝐵 by the bisectors the ray from 𝐶 through 𝐷 and the ray from 𝐵 through 𝐷 that intersect at 𝐷. Use the ruler to measure the perimeter of the triangle 𝐷𝐵𝐶 to the nearest one decimal place. Option (A) 17.6 centimeters, option (B) 17.9 centimeters, option (C) 16.9 centimeters, option (D) 18.4 centimeters. Or is it option (E) 19.1 centimeters?

In this question, we are told to construct a triangle 𝐴𝐵𝐶 of given lengths and then construct the angle bisectors of two of its angles. We know that these will intersect at a point, and we can label this point 𝐷. We need to find this point and then measure the perimeter of triangle 𝐷𝐵𝐶 to the nearest tenth of a centimeter.

To begin, we might be tempted to use the given diagram. However, there might be inaccuracies in the diagram. Instead, we should construct the triangle ourselves to minimize errors. To do this, we need to start with sketching any side of the triangle. Let’s say we draw a line of length nine centimeters as shown. We can then trace a circle of radius six centimeters centered at 𝐶 and a circle of radius five centimeters centered at 𝐵 and label the point of intersection 𝐴. If we connect the vertices with sides as shown, then we have constructed the triangle with the given side lengths.

We now need to bisect the angles at 𝐶 and 𝐵. We can do this by recalling the method for constructing an angle bisector using a pair of compasses and a ruler. We can start by bisecting the angle at 𝐶. We first need to trace a circle centered at 𝐶 that intersects both sides of the angle. We will call these points 𝐸 and 𝐹 as shown. Next, we trace congruent circles centered at 𝐸 and 𝐹 that intersect at a point on the same side as the angle we want to bisect. We will call this point 𝐺 as shown. Finally, we can conclude that the line between 𝐶 and 𝐺 is the bisector of the angle. We will extend this bisector to the opposite side of the triangle.

We now need to follow this same process to bisect the angle at 𝐵. We start by tracing a circle at 𝐵 and labeling the points of intersection between the circle and the sides of the angle 𝐼 and 𝐻 as shown. Next, we trace congruent circles at 𝐼 and 𝐻 and label the point of intersection between the circles 𝐽. We can then conclude that the line between 𝐽 and 𝐵 is the bisector of the angle at 𝐵. We can extend this line to the opposite side of the triangle. We can now label the point of intersection between the bisectors 𝐷.

We now want to estimate the perimeter of triangle 𝐷𝐵𝐶, that is, the sum of the side lengths. If we measure the two unknown side lengths of this triangle with a ruler, we can estimate that 𝐶𝐷 is approximately 5.2 centimeters long and 𝐵𝐷 is approximately 4.2 centimeters long. Adding these lengths together gives us a perimeter of 18.4 centimeters to the nearest tenth of a centimeter, which we can see agrees with option (D).

There is another useful property of bisectors that we can note by using this example. Let’s construct the bisector of angle 𝐴 in the same way we did for angles 𝐵 and 𝐶. Following this process, we can note that the bisector of 𝐴 also intersects the other bisectors at point 𝐷. This result holds true for any triangle. In general, we have that the internal angle bisectors of any triangle intersect at a single point. In other words, they are concurrent. In our final example, we will construct a right trapezoid and then use the angle bisectors to estimate the length of a line segment.

Given that 𝐴𝐵𝐶𝐷 is a trapezoid, use a ruler and a pair of compasses to draw the trapezoid and bisect angle 𝐴 by the bisector the ray from 𝐴 through 𝐷 that intersects the ray from 𝐶 through 𝐵 at 𝑀. Use a ruler to measure the length of the line segment 𝐴𝑀 to the nearest one decimal place. Option (A) 12.0 centimeters, option (B) 10.9 centimeters, option (C) 14.1 centimeters, option (D) 11.6 centimeters. Or is it option (E) 13.0 centimeters?

In this question, we are given a right trapezoid 𝐴𝐵𝐶𝐷 and asked to construct an exact version of this trapezoid using a ruler and a pair of compasses. We then need to find a point 𝑀 by bisecting the angle at 𝐴 and then measure the length of the line segment between 𝐴 and 𝑀 to the nearest tenth of a centimeter. Let’s start by constructing an exact version of the trapezoid using a ruler and a pair of compasses. We can start by measuring a line segment of length 10 centimeters and labeling the endpoints 𝐴 and 𝐵 as shown.

Next, we need to construct a right angle at 𝐵. And we can do this by bisecting a straight angle at 𝐵. We start by extending the line segment beyond 𝐵 and then tracing a circle centered at 𝐵 that intersects the line at 𝐸 and 𝐹 as shown. We then trace congruent circles centered at 𝐸 and 𝐹 that intersect at a point above 𝐵 as shown. We can then note that the line through 𝐵 and this point of intersection bisects the straight angle, so it will be a right angle. If we draw a straight line segment of length 12 centimeters, then we can label the other endpoint 𝐶.

We could follow this process again to construct a right angle at 𝐶. However, we can also note that 𝐴𝐶𝐷 is a triangle. And we can construct this triangle using a ruler and a straightedge. We trace a circle of radius 13 centimeters centered at 𝐴 and a circle of radius five centimeters centered at 𝐶 and label the point of intersection shown 𝐷 to construct the trapezoid.

We now need to bisect the angle at 𝐴. We do this by tracing a circle centered at 𝐴 that intersects the sides of the angle at two points that we will label 𝐺 and 𝐻. We then trace congruent circles centered at 𝐺 and 𝐻 and label the point of intersection on the same side as angle 𝐴 𝐼. We can then conclude that the line between 𝐴 and 𝐼 bisects the angle at 𝐴. We extend this line to intersect the line between 𝐵 and 𝐶 and label the point of intersection 𝑀. If we then measure the length of this line segment to the nearest millimeter, we get 12.0 centimeters, which we can see is option (A).

Let’s now go over the key points from this lesson. We started by defining the bisector of an angle to be the line through the vertex of the angle that splits the angle in half, or in other words into two congruent angles. Next, we proved that the diagonal of a kite that lies between its congruent sides bisects two of the internal angles of a kite. In the same way, we can extend this to both diagonals of a rhombus.

We then use this property to find a construction to bisect any angle by using a pair of compasses and a straightedge. First, we trace a circle centered at the vertex of the angle we want to bisect that intersects the sides of the angle at distinct points that we label 𝐷 and 𝐸, respectively. Next, we trace congruent circles centered at 𝐷 and 𝐸 that intersect at a point 𝐹 on the same side as the angle we want to bisect. Finally, we can conclude that the line between 𝐴 and 𝐹 is the diagonal of a kite between its congruent sides. So it bisects the angle 𝐵𝐴𝐶. Lastly, we saw that the three angle bisectors of the internal angles in any triangle are concurrent. In other words, they all intersect at a single point.