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Lesson Explainer: Geometric Construction: Congruent Angles and Parallel Lines Mathematics

In this explainer, we will learn how to construct an angle to be congruent to a given angle and construct a line to be parallel to a given line.

Congruent angles are an important part of geometry since they are used in constructing congruent shapes and proving geometric properties using angle congruency. The ability to construct an angle congruent to another angle using a compass and straightedge is useful in both of these processes. It is important to note that we want to do this without using a protractor since a protractor will always have some errors in measurement.

To do this, we will use the fact that any two triangles with the same side lengths must have their corresponding angles congruent. This is called the SSS congruency criterion.

Before we duplicate an angle, it is worth noting that if the angle is a zero angle, a straight angle, or a full rotation, then the sides of the angle form either a line or a ray, and these are duplicated by just using the straightedge.

Let’s start with a given ∠𝐴𝐡𝐢 that we want to duplicate.

We are going to do this by constructing a congruent triangle. So, for now, we will assume that m∠𝐴𝐡𝐢<180∘.

We start by tracing a circle centered at 𝐡 that intersects the edges of the angle at two points we will label 𝐢′ and 𝐴′.

We can connect these points with a straight line to construct triangle 𝐴′𝐡𝐢′.

We want to construct a triangle congruent to triangle 𝐴′𝐡𝐢′ so that we duplicate the angle at 𝐡. To do this, let’s start with a ray 𝐸𝐷.

If we set the radius of our compass equal to 𝐴′𝐡 and then trace a circle centered at 𝐸, we can find a point of intersection between the circle and ray that we call 𝐷′. We note that 𝐸𝐷′=𝐴′𝐡 since they are both radii of congruent circles.

We note that we need to trace this arc centered at 𝐸 further than the measure of angle 𝐡. We note that every point on this arc is a distance of 𝐴′𝐡 from 𝐸. In particular, 𝐡𝐢′=𝐴′𝐡, so any point on this circle will give us a side congruent to 𝐡𝐢′.

Let’s now change the radius of the compass to be 𝐴′𝐢′. We can then trace a circle centered at 𝐷′ with this radius to find a point of intersection between the circles; we call this point 𝐹.

We now sketch line segments 𝐷′𝐹 and 𝐸𝐹, and we note that 𝐷′𝐹=𝐴′𝐢′ and 𝐸𝐹=𝐡𝐢′.

Since triangles 𝐴′𝐡𝐢′ and 𝐷′𝐸𝐹 have three congruent sides, we can conclude, using the SSS congruency criterion, that the corresponding angles are congruent. In particular, this tells us that π‘šβˆ π΅=π‘šβˆ πΈ.

It is worth noting that this construction remains unchanged for an obtuse or a right angle at 𝐡. In both cases, we construct the congruent triangle 𝐷′𝐸𝐹 in exactly the same way. For example, for an obtuse angle at 𝐡, we get the following.

We can extend this construction to reflex angles by noting that we can duplicate the smaller angle at the vertex, and this also duplicates the reflex angle. For example, if we want to duplicate the following reflex angle, we instead follow the process to duplicate the smaller angle at 𝐡.

We then see that the reflex angle at 𝐸 is congruent to the reflex angle at 𝐡.

This means we can use this process to duplicate any angle; however, we generally leave out the triangle construction and just use the arcs to find the duplicate angle.

How To: Duplicating an Angle with a Compass and Straightedge

We can construct a congruent angle to any ∠𝐴𝐡𝐢 with a compass and straightedge. First, if ∠𝐴𝐡𝐢 is a zero angle, a straight angle, or a full turn, then we just need to sketch a line and mark an angle with the same measure. Otherwise, we use the following steps:

  1. Sketch a ray 𝐸𝐷; 𝐸 will be the vertex of the congruent angle.
  2. Trace a circle centered at 𝐡 that intersects the edges of the angle at two points we will name 𝐢′ and 𝐴′.
  3. Trace a circle of radius 𝐴′𝐡 centered at 𝐸. Label the point of intersection between the ray and circle 𝐷′.
  4. Set the radius of the compass equal to 𝐴′𝐢′ and then trace a circle centered at 𝐷′. Label the point of intersection between the two circles 𝐹.
  5. We then have that π‘šβˆ π΄π΅πΆ=π‘šβˆ π΄β€²π΅πΆβ€²=π‘šβˆ π·β€²πΈπΉ.

Let’s now see an example of correctly identifying the first step in the construction of a congruent angle with a compass and straightedge.

Example 1: Correctly Describing the First Step to Construct a Congruent Angle

Fill in the blanks: The first step to construct a congruent angle to a given angle after drawing a ray is to draw from the vertex of the first angle, then to draw from the vertex of the second angle.

  1. a straight line, a straight line with a different length
  2. an arc with a suitable radius, a straight line
  3. an arc with a suitable radius, an arc with the same radius
  4. an arc with a suitable radius, an arc with a different radius

Answer

To construct a congruent angle with a compass and straightedge, we want to construct a congruent triangle since the corresponding angles will then be equal. To do this, we first sketch a ray that we will duplicate the angle onto. Then, we trace a circle at the vertex of the angle so that the two sides of the triangle have equal length.

We can then trace a circle of equal radius centered at the endpoint of the ray to get the following.

Finally, we trace a circle of radius 𝐴′𝐢′ centered at 𝐷′ to construct congruent triangles 𝐴′𝐡𝐢′≅𝐷′𝐸𝐹.

Hence, the answer is option C.

The first step to construct a congruent angle to a given angle after drawing a ray is to draw an arc with a suitable radius from the vertex of the first angle, then to draw an arc with the same radius from the vertex of the second angle.

In our next example, we will determine which of a list of different constructions correctly duplicates an angle.

Example 2: Identifying the Construction of Congruent Angles

Which of these two figures shows the steps for constructing a congruent angle?

Answer

Let’s say these constructions are supposed to duplicate ∠𝐷𝐸𝐹.

We recall that we construct a congruent angle with a compass and straightedge by sketching a ray 𝐴𝐺, where 𝐴 will be the vertex of the congruent angle. We then trace a circle centered at 𝐸 that intersects the edges of the angle at two points, 𝐷′ and 𝐹′. We trace a congruent circle centered at 𝐴 and label the point of intersection 𝐡.

Finally, we set the radius of the compass equal to 𝐷′𝐹′ and trace a circle of this radius centered at 𝐡. The point of intersection of these circles is point 𝐢, and we have that π‘šβˆ π·πΈπΉ=π‘šβˆ π΅π΄πΆ.

We can see that this is the construction in Figure I.

For due diligence, we should also check if the construction in Figure II duplicates an angle. We note that the two congruent circles are traced with centers at 𝐡 and 𝐢. This means that we have the following congruent line segments.

This does not duplicate the angle. In fact, we can note that this bisects the angle by adding in the following diagonal and point.

We see that triangles 𝐴𝐺𝐢 and 𝐴𝐺𝐡 are congruent by the SSS criterion, so π‘šβˆ πΊπ΄πΆ=π‘šβˆ πΊπ΄π΅.

Hence, only Figure I shows the construction of a congruent angle.

Before we move on to constructing parallel lines, we first need to recall a property about transversals of parallel lines. Let’s first see a specific example of this property.

Example 3: Using Congruence of Corresponding Angles to Identify Geometric Relationships

True or False: ⃖⃗𝐹𝐡 is parallel to οƒͺ𝐢𝐸.

Answer

We begin by recalling that if the corresponding angles of a transversal of two lines are congruent, then the lines are parallel. Since we are given that π‘šβˆ π΅π΄πΆ=π‘šβˆ πΈπΆπ·=55∘ and these are corresponding angles, we can conclude that ⃖⃗𝐹𝐡 is parallel to οƒͺ𝐢𝐸.

The property in the previous example holds for any corresponding congruent angles.

In our next example, we will use the construction of congruent angles and the property of corresponding congruent angles in a transversal making the line parallel to determine a geometric relationship.

Example 4: Determining a Geometric Property by Constructing a Congruent Angle

Draw △𝐴𝐡𝐢 where 𝐴𝐡=3cm, 𝐡𝐢=4cm, and 𝐴𝐢=5cm. Point 𝐷 lies on οƒͺ𝐡𝐢 such that 𝐷 is not on line segment 𝐡𝐢. Draw ∠𝐷𝐢𝐸 congruent to ∠𝐢𝐡𝐴, where 𝐸 is on the upper side of 𝐡𝐢. Which of the following is true?

  1. π‘šβˆ π΄=π‘šβˆ πΈπΆπ΄
  2. π‘šβˆ π΄=π‘šβˆ π΅πΆπ΄
  3. π‘šβˆ π΄=π‘šβˆ πΈπΆπ·
  4. π‘šβˆ π΅=π‘šβˆ πΈπΆπ΄

Answer

We want to start by sketching △𝐴𝐡𝐢 and there are a few ways of doing this. One way is to note that 3, 4, 5 is a Pythagorean triple, so △𝐴𝐡𝐢 is a right triangle. However, it is not necessary to notice this to sketch the triangle.

We can start by drawing a 3 cm long line segment 𝐴𝐡. Next, we trace a 4 cm circle centered at 𝐡 and a 5 cm circle centered at 𝐴; then, the point of intersection between these circles is 𝐢.

We now need to extend 𝐡𝐢 to make it a ray and then find the point on this ray, not on 𝐡𝐢, such that ∠𝐷𝐢𝐸 is congruent to ∠𝐢𝐡𝐴. We will do this by using the construction for congruent angles.

First, let’s clear the previous construction to make the process clearer. We will start with only triangle 𝐴𝐡𝐢 where we will mark ∠𝐢𝐡𝐴.

We now trace a circle at 𝐡 that intersects the sides of the triangle as shown.

It is worth noting that the radius of this circle does not matter; however, it is easiest to choose a radius smaller that 1.5 cm. The choice of the radius will change where we sketch our point 𝐷.

We then trace a congruent circle at 𝐢 to get a possible point 𝐷 as shown.

We now set the radius of our compass to be equal to the length of the line segment between the two points of intersection on the edges of the angle we want to duplicate.

We trace a circle of this radius centered at 𝐷 and label the point of intersection between the circles 𝐸.

Adding labels for the two points of intersection on the edges of the angle we wanted duplicated, we have that triangles 𝐴′𝐡𝐢′ and 𝐸𝐢𝐷 are congruent. Hence, ∠𝐸𝐢𝐷 is congruent to ∠𝐢𝐡𝐴 (i.e., ∠𝐡).

We now recall that if a transversal of two lines makes the same angle with both lines, then the lines are parallel. Since π‘šβˆ πΈπΆπ·=π‘šβˆ π΅ and these are the angles that the transversal οƒͺ𝐡𝐢 makes with 𝐢𝐸 and 𝐴𝐡, then 𝐢𝐸 is parallel to 𝐴𝐡.

Now, since 𝐴𝐢 is a transversal to a pair of parallel lines, we recall that the alternate interior angles ∠𝐴 and ∠𝐸𝐢𝐴 must be congruent.

Hence, π‘šβˆ π΄=π‘šβˆ πΈπΆπ΄, which is option A.

In the previous example, we used a property of parallel lines and our construction of congruent angles to show that two lines are parallel. We can generalize this process to construct a line parallel to any given line through a point.

For example, let’s say we want to construct a line parallel to ⃖⃗𝐴𝐡 through 𝐢.

We sketch the line ⃖⃗𝐴𝐢 and the ∠𝐡𝐴𝐢.

We can duplicate ∠𝐡𝐴𝐢 at 𝐢. First, we trace a circle centered at 𝐴 and then trace a congruent circle centered at 𝐢, labeling the points as shown.

We then set the radius of the compass to 𝐡′𝐢′ and trace a circle of this radius centered at 𝐷. We can label the point of intersection between the circles 𝐸, and we have that ∠𝐡𝐴𝐢 is congruent to ∠𝐸𝐢𝐷.

Sketching the line ⃖⃗𝐢𝐸, we can then note that since the corresponding angles are equal, we must have that ⃖⃗𝐢𝐸 is parallel to ⃖⃗𝐴𝐡.

Hence, we can construct a line parallel to ⃖⃗𝐴𝐡 through 𝐢 by constructing an angle congruent to ∠𝐡𝐴𝐢 at 𝐢.

In our final example, we will construct a line parallel to another line to show a useful geometric property.

Example 5: Finding the Measure of an Angle by Constructing a Parallel Line

In the following figure, ⃖⃗𝐴𝐡⫽⃖⃗𝐢𝐷, while ⃖⃗𝐸𝐹 cuts ⃖⃗𝐴𝐡 and ⃖⃗𝐢𝐷 at 𝑋 and π‘Œ respectively. Draw straight line ⃖⃗𝑀𝑁, where ⃖⃗𝑀𝑁⫽⃖⃗𝐸𝐹 and cuts ⃖⃗𝐴𝐡 and ⃖⃗𝐢𝐷 at 𝑂 and 𝑃, respectively, on the right of ⃖⃗𝐸𝐹. Find π‘šβˆ π‘‚π‘ƒπ‘Œ.

Answer

To sketch ⃖⃗𝑀𝑁, we first need to choose where we want this line to intersect one of the given lines. We will choose a point 𝑂 on ⃖⃗𝑀𝑁 and then construct an angle congruent to βˆ π΄π‘‹πΈ at 𝑂.

We start by tracing a circle centered at 𝑋 that intersects the edges of the angle and then we trace a congruent circle centered at 𝑂, labeling the points of intersection as shown.

We then set the radius of the compass equal to 𝐴′𝐸′ and trace a circle of this radius centered at 𝑋′. We call the point of intersection between the arcs 𝑀, and we have that βˆ π‘€π‘‚π‘‹β€² is congruent to βˆ πΈπ‘‹π΄.

Hence, both of these angles have a measure of 80∘. We can also label the point of intersection between ⃖⃗𝑀𝑂 and ⃖⃗𝐢𝐷 as 𝑃. This allows us to mark βˆ π‘‚π‘ƒπ‘Œ on the diagram.

We can see that ⃖⃗𝑂𝑃 is a transversal of the parallel lines ⃖⃗𝐴𝐡 and ⃖⃗𝐢𝐷. This means that the corresponding angles must be congruent. Hence, π‘šβˆ π‘‚π‘ƒπ‘Œ=π‘šβˆ π‘€π‘‚π‘‹=80.∘

In the previous example, quadrilateral π‘‚π‘‹π‘Œπ‘ƒ has opposite sides that are parallel, so it is a parallelogram. Let’s note two properties of parallelograms.

First, as we have seen in the previous example, the diagonally opposite angles in a parallelogram are congruent.

Second, we recall that a straight angle has a measure of 180∘. So, we can see that π‘šβˆ π‘‹π‘‚π‘ƒ=180βˆ’80=100∘∘∘, and we know this is congruent to βˆ π‘ƒπ‘Œπ‘‹.

We can use this process to show that these properties hold true for any parallelogram.

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

Key Points

  • We can construct a congruent angle to any ∠𝐴𝐡𝐢 with a compass and straightedge. First, if ∠𝐴𝐡𝐢 is a zero angle, a straight angle, or a full turn, then we just need to sketch a line and mark an angle with the same measure. Otherwise, we use the following steps:
    1. Sketch a ray 𝐸𝐷; 𝐸 will be the vertex of the congruent angle.
    2. Trace a circle centered at 𝐡 that intersects the edges of the angle at two points that we will name 𝐢′ and 𝐴′.
    3. Trace a circle of radius 𝐴′𝐡 centered at 𝐸. Label the point of intersection between the ray and circle 𝐷′.
    4. Set the radius of the compass equal to 𝐴′𝐢′ and then trace a circle centered at 𝐷′. Label the point of intersection between the two circles 𝐹.
    5. We then have that m∠𝐴𝐡𝐢=π‘šβˆ π΄β€²π΅πΆβ€²=π‘šβˆ π·β€²πΈπΉ.
  • We can construct a line parallel to ⃖⃗𝐴𝐡 through point 𝐢 by constructing an angle congruent to ∠𝐡𝐴𝐢 at 𝐢.

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