### Video Transcript

In this video, we will learn how to
find missing lengths in a triangle containing two or three parallel lines using
proportionality. So letβs start with this
triangle. Letβs say this triangle has three
different angle measures. And then this triangle is cut by a
line that is parallel to one of the side lengths. We can label our triangle
π΄π΅πΆ. And weβll let the parallel segment
be segment π·πΈ.

We started with the larger
triangle, triangle π΄π΅πΆ. But now because of this line in the
middle, we have a second triangle, the smaller triangle, triangle π΄π·πΈ. And if we want to compare triangle
π΄π΅πΆ to triangle π΄π·πΈ, we need to be able to say something about its side
lengths or about its angles.

To do this, letβs extend our
parallel lines and the line segment π΄π΅. This should remind us that when two
parallel lines are crossed by a transversal, corresponding angles are equal. Line segment π·πΈ and line segment
π΅πΆ are parallel, which means line segment π΄π΅ could be considered a transversal
of these two parallel lines, which means that angle π· is a corresponding angle with
angle π΅. And these angles will be equal.

For the same reasons, angle πΈ is a
corresponding angle to angle πΆ. And therefore, these two angles
will be equal. Both of these triangles share angle
π΄. So we can say that angle π΄ is
equal to angle π΄, angle π΅ is equal to angle π·, and angle πΆ is equal to angle
πΈ.

When two triangles have three
congruent angles, we can say that the two triangles are similar. This means they are the same shape,
but not the same size. And in similar triangles,
corresponding sides are always proportional. That is, they always occur in the
same ratio.

Letβs look at the corresponding
side lengths for these two triangles. In our larger triangle, we have
side length π΄π΅, which corresponds to the side length π΄π· in the smaller
triangle. This ratio will be equal to side
length π΄πΆ in our larger triangle over side length π΄πΈ in our smaller
triangle. And for our final sides, side
length π΅πΆ in the larger triangle corresponds to side length π·πΈ in the smaller
triangle.

For these ratios, the numerator is
a side length from the larger triangle and the denominator is the corresponding side
length from the small triangle. And in order for our proportion to
hold true, we have to maintain this pattern for the rest of the ratios, the side
length from the larger triangle in the numerator and the corresponding smaller side
length in the denominator.

However, there is another way we
can write these proportions. If we take the larger side length
π΄π΅ and the other larger side length π΅πΆ, we can still set up a ratio. This time, in our numerator on the
other side, weβll need the corresponding side length to π΄π΅, which is π΄π·. And then the denominator, weβll
need the corresponding side to side length π΅πΆ, which in our case is π·πΈ. In this case, we had the ratio of
two of the larger side lengths set equal to the ratio of corresponding side lengths
for the smaller triangle. This means we can line up
corresponding side lengths this way or this way, as long as we are consistent for
each of the fractions.

What weβve shown here can be
summarized in a theorem about triangles. And that is the side splitter
theorem. It says if a line is parallel to a
side of a triangle and the line intersects the other two sides, then the line
divides those sides proportionally. This is true because those parallel
lines create two similar triangles. We should also note that that
parallel line that intersects the other two sides can happen anywhere along the
triangle and between any of the two sides, as long as itβs parallel to the third
side. At this point, weβre ready to
consider some examples.

Using the diagram, which of the
following is equal to π΄π΅ over π΄π·? (A) π΄π΅ over π·π΅, (B) π΄πΆ over
π΄πΈ, (C) π΄πΆ over πΈπΆ, (D) π΄π· over π·π΅, or (E) π΄πΈ over πΈπΆ.

In our diagram, line segment πΈπ·
is parallel to line segment πΆπ΅. Because of that, we have two
similar triangles. We can say that triangle π΄πΈπ· is
similar to triangle π΄πΆπ΅. And in similar triangles,
corresponding side lengths are proportional. Weβre interested in the ratio π΄π΅
over π΄π·. π΄π΅ represents a larger side
length, and π΄π· is the corresponding smaller side length. This means that we are looking for
the ratio that has a larger side length and the corresponding smaller side
length.

Option (A) has π΄π΅ corresponding
to π·π΅. But π·π΅ is not part of the smaller
triangle, which means option (A) cannot work. Option (B) has side length π΄πΆ,
which is part of our larger triangle, and then the distance from π΄ to πΈ. π΄ to πΈ is the corresponding
smaller side length from π΄πΆ. This is an equal ratio. But letβs check the others just in
case.

Again, we have the side length
π΄πΆ, but the denominator is πΈπΆ. And πΈπΆ is not part of the smaller
triangle, which makes this an invalid ratio. What about option (D)? π΄π· is a smaller side length, and
then π·π΅, which is not part of our similar triangles. And we see that again π΄πΈ is a
smaller triangle, but πΈπΆ is not part of any of our similar triangles. The only ratio that is equal to
π΄π΅ over π΄π· in this list is π΄πΆ over π΄πΈ.

In our next example, weβll do
something very similar, only this time weβll need to find the value of a missing
side length.

Find the value of π₯.

When we look at our diagram, we see
a larger triangle that is cut by the segment π·πΈ. And this line segment π·πΈ is
parallel to one of the side lengths π΅πΆ. And when a triangle is cut by a
line segment thatβs parallel to one of its side lengths, two similar triangles are
created. So we can say that if weβre given
line segment π·πΈ is parallel to π΅πΆ, the smaller triangle, triangle π΄π·πΈ, is
similar to the larger triangle, triangle π΄π΅πΆ. So we can say that π΄π· over π΄π΅
is equal to π·πΈ over π΅πΆ. This is because in similar
triangles corresponding side lengths are proportional.

Side length π΄π· measures 10, but
what about π΄π΅? And this is where we need to be
very careful. Side length π΄π΅ is not equal to
11. Side length π΄π΅ is the full
distance from vertex π΄ to vertex π΅, which is 21. So π΄π· over π΄π΅ will be equal to
10 over 21. We know that side length π·πΈ is
equal to 10 as well. So we now have a proportion that
says 10 over 21 is equal to 10 over π΅πΆ. And that means π΅πΆ must be equal
to 21 so that these side lengths stay in proportion. Since side length π΅πΆ equals 21,
our missing π₯-value is 21.

Our next example is another case
where we need to find a missing length in a triangle.

In a figure, segments ππ and π΅πΆ
are parallel. If π΄π equals 18, ππ΅ equals 24,
and π΄π equals 27, what is the length of line segment ππΆ?

The first thing we can do is take
the information in the question and add it to our figure. We know that π΄π equals 18, ππ΅
equals 24, and π΄π equals 27. Our missing length is from π to
πΆ, here. Letβs call it π. Before we start solving for π,
letβs see what we recognize in the figure. The line segment ππ cuts our
triangle π΄π΅πΆ and is parallel to the line segment π΅πΆ. We know that if a line is parallel
to a side of a triangle and the line intersects the other two sides, then the line
divides those sides proportionally. That is, it creates two similar
triangles. So we can start out by saying the
smaller triangle, triangle π΄ππ, is similar to the larger triangle, triangle
π΄π΅πΆ. And since this is true,
corresponding side lengths will be proportional. If we set up a ratio for our
smaller triangle side lengths, we can say that π΄π over π΄π will be equal to π΄π΅
over π΄πΆ.

In order to solve for our missing
length, weβll need to carefully plug the information we know into this ratio. First of all, π΄π is equal to 18
and π΄π equals 27. But we need to be careful with
π΄π΅. To find π΄π΅, we need to add 18 and
24, which gives us 42. But what about the length π΄πΆ? Well, π΄πΆ is made up of π΄π and
ππΆ. π΄π measures 27, and ππΆ measures
π, a missing value. So we plug in 27 plus π plus our
missing value for the length of line segment π΄πΆ.

Once we get to this point, we can
use cross multiplication to solve for our missing variable. 18 times 27 plus π is equal to 27
times 42. 27 times 42 is 1134. From here, we can divide both sides
by 18, which gives us 27 plus π is equal to 63. And when we subtract 27 from both
sides, our missing side length π is equal to 36. If we plug that back into our
diagram, we show that ππΆ equals 36.

In our final example, weβll look at
a figure that includes a parallelogram and a triangle.

πΉπ·πΈπΆ is a parallelogram, where
πΉ and π· are the midpoints of line segment π΄π΅ and line segment π΄πΆ,
respectively, and πΆπΈ equals six centimeters. Determine the length of π΅πΆ.

Letβs take the information weβre
given and add that in to our diagram. If πΉπ·πΈπΆ is a parallelogram, the
opposite sides are parallel, meaning πΉπ· is parallel to πΆπΈ and πΉπΆ is parallel
to π·πΈ. We also know that πΆπΈ measures six
centimeters. But based on our parallelogram
properties, we also know that opposite side lengths will be equal. And that means that π·πΉ must also
measure six centimeters.

But our missing side length is
π΅πΆ. And so weβll need to be able to say
something else here. If πΆπΈ is parallel to πΉπ·, we can
also say that π΅πΈ is parallel to πΉπ·. And if we can say that line segment
π΅πΆ is parallel to line segment πΉπ·, then we have a line that is parallel to a
side of our triangle. And that line intersects the other
two sides, which means the line segment πΉπ· creates two similar triangles. The smaller triangle, triangle
π΄π·πΉ, is similar to the larger triangle, triangle π΄πΆπ΅. And in similar triangles,
corresponding side lengths are proportional. The side length π΄π· on the smaller
triangle corresponds to the side length π΄πΆ on the larger triangle. And that will have to be equal to
the smaller triangleβs line segment πΉπ· over the larger triangleβs line segment
π΅πΆ.

If we try to plug in the
information we know, we only end up filling in the side length πΉπ·, which is six
centimeters. But we can think carefully about
this midpoint π·. The midpoint π· divides line
segment π΄πΆ in half. And so we can say that π΄πΆ is
equal to two times π΄π·. What weβre saying is the ratio of
the smaller triangle to the larger triangle would be one-half since the larger
triangle is always two times greater than the smaller triangle. And if the side lengths of the
larger triangle is two times that of the smaller triangle and πΉπ· is equal to six
centimeters, we know that π΅πΆ will have to be equal to 12 centimeters, as 12 is
twice six.

Before we finish, letβs just review
some key points from this video. If a line is parallel to a side of
a triangle and the line intersects the other two sides, then that line creates two
similar triangles. And in similar triangles,
corresponding side lengths are proportional. For this example, triangle π΄π·πΉ
would be similar to triangle π΄π΅πΆ. Therefore, the side length π΄π·
over the side length π΄π΅ will be equal to the side length π΄πΉ over the side length
π΄πΆ.