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.