### Video Transcript

Geometric Construction: Congruent Angles and Parallel Lines

In this lesson, we will learn how to construct an angle to be congruent to a given
angle using only a compass and a straight edge. We will then learn how to apply this process to construct a line parallel to another
line through a given point.

Congruent angles are an important part of geometry since theyβre used in constructing
congruent shapes and in proving many geometric properties using congruency. The process of constructing an angle congruent to another angle is useful in both of
these processes. It is important to note that we want to do this without using a protractor, since
protractors will have errors in measurements, whereas a geometric construction will
theoretically have no errors at all.

Letβs start with an angle π΄π΅πΆ that we want to duplicate. In this case, we can see that this angle is acute. To duplicate this angle, we want to apply the side-side-side congruency criterion for
triangles. In other words, we are going to instead construct a triangle with angle π΄π΅πΆ as an
internal angle and then duplicate this triangle, which will also duplicate the
angle.

Letβs start by sketching a circle centered at π΅, which intersects the sides of the
angle as shown. We only draw the arcs of this circle for simplicity. We can label the points of intersection π΄ prime and πΆ prime. We can now see that triangle π΄ prime π΅πΆ prime has an internal angle with the same
measure as angle π΄π΅πΆ. We want to duplicate this triangle.

To duplicate this triangle, and hence the angle, we will start by sketching a ray in
the plane. It is worth noting that we can sketch the ray in any orientation. It does not have to be in the same direction as π΅πΆ. However, for simplicity, we will sketch our ray in a similar direction as shown. We now trace a circle of radius πΆ prime π΅ centered at πΈ. We can now see that line segments πΈπ· prime and π΅πΆ prime are the same length. In fact, the distance between any point on this circle and πΈ is the same length as
these lines.

Letβs now change the radius of our compass to be the length π΄ prime πΆ prime and
trace a circle of this radius centered at π· prime. We can call the point of intersection between the circles πΉ as shown. We can now note that we have duplicated all of the sides of triangle π΄ prime π΅πΆ
prime as a new triangle πΉπΈπ· prime. In particular, by the SSS criterion, this means that we have duplicated the angle at
π΅.

We can write down the steps we followed in this process to describe how we can
duplicate an angle π΄π΅πΆ. We start by sketching a ray from πΈ through π· anywhere on the plane, where our
construction will duplicate the angle such that πΈ is the vertex of the congruent
angle. Next, we trace a circle centered at π΅ that intersects the sides of the angle. We will label these points of intersection π΄ prime and πΆ prime. Then, we trace a congruent circle of radius πΆ prime π΅ centered at πΈ. We label the point of intersection between the ray from πΈ through π· and this circle
π· prime. Finally, we trace a circle of radius π΄ prime πΆ prime centered at π· prime and label
the point of intersection between the two circles πΉ. Then, angle π΄π΅πΆ is congruent to angle πΉπΈπ· prime.

There are a few things worth noting about this process. First, we started by assuming that our angle was acute. So we can ask if this process works for angles which are not acute. This process does indeed work for any angle. However, it is worth noting that if the angle we want to duplicate is a straight
angle, a full turn, or the zero angle, then it is easier to just sketch a ray or
line. We can then guarantee that the angle is duplicated.

We can show that this construction works to duplicate any reflex angle by noting that
we can just use this construction to duplicate the acute angle. Thus, this will also duplicate the reflex angle.

There are two more cases we need to consider, that is, right angles and obtuse
angles. The case for a right angle is the same as the case for an acute angle. So we will not show this in the video. Instead, letβs check the process for the obtuse angle shown.

First, we start by sketching any ray we will call πΈπ· anywhere in the plane. Second, we trace a circle centered at π΅ that intersects both sides of the angle. We call these points of intersection π΄ prime and πΆ prime. For simplicity, we only sketch the two arcs as shown. Third, we trace a circle of radius πΆ prime π΅ centered at πΈ and label the point of
intersection between this circle and the ray π· prime. We need to sketch this arc to have larger measure than the angle we want to
duplicate. Finally, we trace a circle of radius π΄ prime πΆ prime centered at π· prime and label
the point of intersection between the two circles πΉ. We can see that the triangles π΄ prime π΅πΆ prime and πΉπΈπ· prime are congruent by
the side-side-side criterion. Hence, the angles at π΅ and πΈ have equal measure.

Letβs now see an example of identifying the correct construction of duplicating an
angle.

Which of these two figures shows the steps for constructing a congruent angle? Figure I or figure II.

In this question, weβre given two figures and we need to determine which of the
figures demonstrates the steps of duplicating an angle. We note that constructing means with a compass and a straight edge.

Letβs start by saying that these constructions are supposed to duplicate angle
π·πΈπΉ. We can then recall that to duplicate this angle, we need to start by drawing a ray in
the plane, say the ray from π΄ through πΊ, where π΄ will be the vertex of the angle
we duplicate. We can see that both constructions have such a ray if we add a point πΊ to each
figure as shown.

The next step in our construction is to trace a circle centered at πΈ that intersects
the sides of the angle we want to duplicate at two points we will label π· prime and
πΉ prime. After this, we need to trace a congruent circle centered at π΄. In the diagrams, we only sketch an arc of the circle to keep the construction
clean. We call the point of intersection between the ray and the circle π΅. The final step in the construction is to trace a circle of radius π· prime πΉ prime
centered at π΅. We can call the point of intersection of the two circles πΆ as shown.

We can conclude that the angles are congruent because triangle πΉ prime πΈπ· prime is
congruent to triangle πΆπ΄π΅ by the side-side-side criterion. We see that this is only the case in the first figure.

For due diligence, we can check what the second construction gives us. We see that we have the arcs of two congruent circles centered at πΆ and π΅. This gives us the following pairs of lines of the same length, since they are radii
of the congruent circles. If we connect π΄ to the point of intersection of these circles, π», as shown, then we
can note that we have two congruent triangles. This means that all of the corresponding angles of the two triangles must be
congruent. We can then note that this means that this gives us the angle bisector. Hence, the answer is that only figure I shows the steps for constructing a congruent
angle.

Before we move on to constructing parallel lines, there is an important property of
parallel lines we need to recall by looking at an example.

True or False: The line between πΉ and π΅ is parallel to the ray from πΆ through
πΈ.

In this question, we want to determine if a given line and ray are parallel by using
a given figure. To do this, we can start by highlighting the line and ray in the question on the
diagram as shown. We can then note that the line between π΄ and π· is a transversal of these two lines
and the corresponding angles of the transversal are both congruent since they have
the same measure. They both measure 55 degrees.

We can then recall that if the corresponding angles of a transversal of a pair of
lines are congruent, then the lines themselves are parallel. So the answer must be true. The line between πΉ and π΅ is parallel to the ray from πΆ through πΈ.

In our next example, we will use this property and our construction of congruent
angles to show a geometric property.

Draw triangle π΄π΅πΆ where π΄π΅ equals three centimeters, π΅πΆ equals four
centimeters, and π΄πΆ equals five centimeters. Point π· lies on the ray from π΅ through πΆ such that π· is not on line segment
π΅πΆ. Draw angle π·πΆπΈ congruent to angle πΆπ΅π΄, where πΈ is on the upper side of line
segment π΅πΆ. Which of the following is true? Option (A) the measure of angle π΄ equals the measure of angle πΈπΆπ΄. Option (B) the measure of angle π΄ is equal to the measure of angle π΅πΆπ΄. Option (C), the measure of angle π΄ equals the measure of angle πΈπΆπ·. Or is it option (D) the measure of angle π΅ is equal to the measure of angle
πΈπΆπ΄?

In this question, weβre given a construction and we need to determine which of five
given options is correct in the construction. We need to start by sketching triangle π΄π΅πΆ by using the given side lengths of the
triangle. We could do this by noting that the lengths of three, four, and five centimeters make
a Pythagorean triple. This is enough to know that it is a right triangle with a hypotenuse of five
centimeters by using the side-side-side congruency criterion.

However, it is not necessary to notice this to construct this triangle. In general, we can construct a triangle from its lengths by using circles. We start by drawing one of its sides. Letβs say π΄π΅, which has a length of three centimeters. We can then trace a circle of radius four centimeters centered at π΅ and a circle of
radius five centimeters centered at point π΄. Either point of intersection between these circles can be our point πΆ. We choose πΆ as shown. We note that triangle π΄π΅πΆ has the desired lengths.

We now want to extend the line segment π΅πΆ to be the ray from π΅ through πΆ so that
we can find the point π· on this ray to make the angle π·πΆπΈ congruent to the angle
πΆπ΅π΄. To do this, letβs start by highlighting the angle πΆπ΅π΄ that we want to be congruent
to the angle π·πΆπΈ. We know that this is a right angle since this is a Pythagorean triple. However, it is not necessary to use this to answer the question.

Instead, we can duplicate this angle of vertex πΆ by using our construction for
duplicating an angle at a point. To duplicate this angle, we start by tracing a circle at π΅ with radius less than
three centimeters. We call the points of intersection with the sides of the angle π΄ prime and πΆ
prime. It is worth noting that it is easier to choose a radius below 1.5 centimeters.

Next, we trace a circle of the same radius centered at πΆ. We label the point of intersection between the ray from π΅ through πΆ and the circle
that is above πΆ as point π·. Now, we can duplicate the angle shown at πΆ by tracing a circle of radius π΄ prime πΆ
prime centered at π· and labeling the point of intersection between the circles πΈ
as shown. We then know that angle π·πΆπΈ is congruent to angle πΆπ΅π΄. We can also see that line π΅πΆ is a transversal of the lines πΆπΈ and π΄π΅. And this transversal makes the same angle with both lines. We can recall that this means that the two lines must be parallel.

We can then note that the line between π΄ and πΆ is also a transversal between these
parallel lines. So the alternating angles it makes with each line must be congruent. This gives us that angle πΈπΆπ΄ must have the same measure as angle π΄, which we can
see is option (A).

We can combine our construction of a congruent angle with the property that two lines
are parallel if they have corresponding congruent angles in a transversal to
construct a line parallel to any other line through any point. For instance, letβs say we want to construct a line parallel to π΄π΅ through the
point πΆ that is not on this line as shown.

To do this, we first need a transversal so that we can show that the transversal will
have congruent corresponding angles with our pair of lines. We can sketch the line between π΄ and πΆ to be this transversal. We can see that the angle that line π΄πΆ makes with line π΄π΅ is angle π΅π΄πΆ as
shown. We want to duplicate this angle at πΆ. We duplicate this angle by tracing a circle at π΄ that intersects the sides of the
angle at πΆ prime and π΅ prime as shown. We then trace a congruent circle centered at πΆ and call the point of intersection
between the circle and the line π·.

Now, we trace a circle of radius π΅ prime πΆ prime centered at π· and call the point
of intersection between the circles πΈ as shown. We then know that angle π΅π΄πΆ is congruent to angle π·πΆπΈ. Since the corresponding angles in this pair of parallel lines cut by a transversal
have equal measure, we can conclude that the lines must be parallel.

In our final example, we will use this idea of constructing a line parallel to
another line to show a useful geometric property.

In the following figure, the line between π΄ and π΅ is parallel to the line between
πΆ and π·, while the line between πΈ and πΉ cuts the line between π΄ and π΅ and the
line between πΆ and π· at π and π, respectively. Draw straight line ππ, where the line between π and π is parallel to the line
between πΈ and πΉ and cuts the line between π΄ and π΅ and the line between πΆ and π·
at π and π, respectively, on the right side of the line between πΈ and πΉ. Find the measure of the angle πππ.

In this question, weβre given a lot of information about a given figure. We can start by adding the extra information onto the figure. First, we are told that the lines between π΄ and π΅ and πΆ and π· are parallel. This makes the line between πΈ and πΉ a transversal of this pair of parallel
lines.

Next, weβre told that we need to construct a line between two points, π and π, on
the figure, where this line is parallel to the line between πΈ and πΉ. And it intersects the line between π΄ and π΅ at π and the line between πΆ and π· at
π and is on the right of the line between πΈ and πΉ.

To construct this line, we first need to choose a point of intersection between the
line and one of the lines π΄π΅ and πΆπ·. Letβs choose the point π as shown. We need to make sure that this point is on the line between π΄ and π΅ and that this
point is to the right of the line between πΈ and πΉ. We can then recall that we can construct a line parallel to another line through a
point by duplicating the angle at this point. So we will duplicate angle π΄ππΈ at the point π.

To do this, we first need to trace a circle centered at π that intersects the two
sides of the angle as shown. We will call these points of intersection π΄ prime and πΈ prime. Next, we trace a congruent circle centered at π. And we will call the point of intersection between the circle and the line that is to
the left of π π prime as shown. We now trace a circle of radius π΄ prime πΈ prime centered at π prime and call the
point of intersection between the circle shown π.

We can now note that triangle π΄ prime ππΈ prime is congruent to triangle π prime
ππ by the side-side-side criterion. Hence, angle π΄ prime ππΈ prime is congruent to angle π prime ππ. We can add to the diagram that these angles both have measure 80 degrees. We can also call the point of intersection between line ππ and line πΆπ· π as
shown. And we can also add a point onto the line called π. We can also note that the line between π΄ and π΅ is a transversal of the lines
between π and π and π and π with congruent corresponding angles. So they must be parallel.

We want to find the measure of angle πππ. And we can mark this angle on the diagram as shown. We can note that the line between π and π is a transversal of parallel lines. So the corresponding angles must be congruent. Hence, we can conclude that the measure of angle πππ must be equal to 80
degrees.

In the previous example, we can actually note many more useful properties of this
type of construction. First, we can note that quadrilateral ππππ has opposite sides parallel. So it is a parallelogram. We also know that diagonally opposite angles in a parallelogram are congruent. So we can note that the measure of angle πππ must also be 80 degrees. Similarly, we can note that a straight angle has measure 180 degrees. We can use this to find that the measure of angle πππ is 100 degrees and that the
diagonally opposite angle πππ must have the same measure. We can use this process to prove these properties will hold true for any
parallelogram.

Letβs now go over the key points of this lesson. First, we saw that we can duplicate an angle with a compass and a straight edge. In particular, if the angle we want to duplicate is a zero angle, straight angle, or
full turn, then we only need to use the straight edge to duplicate the angle.

However, if we want to duplicate a different angle, then we need to follow four
steps. The first step is to sketch a ray anywhere in the plane. Letβs say we sketch a ray from πΈ through π· as shown. Second, we need to trace a circle centered at point π΅ such that it intersects both
sides of the angle. We will label these points of intersection π΄ prime and πΆ prime as shown. Third, we trace a congruent circle centered at point πΈ and label the point of
intersection between the circle and the ray π· prime. Finally, we trace a circle of radius π΄ prime πΆ prime centered at π· prime and call
the point of intersection between the two circles πΈ prime.

We can note that triangle π΄ prime π΅πΆ prime and triangle π· prime πΈπΈ prime are
congruent by the side-side-side congruency criterion. So the measure of angle π΄π΅πΆ is equal to the measure of angle π· prime πΈπΈ
prime.

Finally, we saw that we can apply this construction process to construct a line
parallel to another line through a point. In particular, if we want to construct a line parallel to a given line between π΄ and
π΅ through a given point πΆ, we can do this by duplicating the angle π΅π΄πΆ at the
point πΆ.