# Video: Parallel Lines and Transversals: Proportional Parts

In this video, we will use the properties of parallel lines and transversals to find a missing length of a line segment in a transverse line cut by parallel lines.

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### Video Transcript

In this video, we will use properties of parallel lines to find a missing length of a line segment in a transversal cut by parallel lines. To do that, letβs consider proportional parts of parallel lines. Itβs a property of parallel lines that says if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. Letβs see what this looks like. Here are three parallel lines, and here are two transversals. Remember, a transversal just means a line that crosses at least two other lines.

In our image, these two lines are transversals. In the figure, we see that the three parallel lines intersected by the two transversals create four different line segments. And if the lengths of those line segments β we have labeled here π, π, π, π β based on this principle, these values will be proportional. And so we can say that π over π will be equal to π over π. Or thereβs one other way we can write this proportion. π over π is equal to π over π. In both cases, when we take the cross product, weβll have π times π is equal to π times π or π times π.

At this point, we should recognize that this property will also be true inside of polygons. To see that, we can modify this figure. Weβve modified our figure so that it is now a trapezoid π΄π΅πΆπ·. And this trapezoid is cut by the line segment πΈπΉ. And πΈπΉ is parallel to the line segment of the trapezoid π΄π· and π΅πΆ. And so we can say that the line segment πΈπΉ cuts this trapezoid into proportional line segments, where π΄πΈ is proportional to πΈπ΅ and π·πΉ is proportional to πΉπΆ.

Letβs consider one more polygon. If we have the larger triangle π΄π΅πΆ being cut by two parallel lines πΈπΉ and πΊπ», where the parallel lines are parallel to the side of the larger triangle πΆπ΅, we can say that π over π will be equal to π over π, which would be equal to π over π. We have proportional side lengths created by these parallel lines.

As an extension of this, thereβs one more property we need to consider. And that is congruent segments on transversals. If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. Hereβs our three parallel lines, and hereβs a transversal. If we know that the segments cut off by these three parallel lines are congruent, then for any other transversal between these three parallel lines, their segments cut off by the parallel lines will be congruent to each other.

Be careful here though. Weβre saying that π will be congruent to π and π will be congruent to π. This does not mean that this line segment with a length π is equal to the line segment with a length π. Weβre saying that they will be congruent on their transversal, not between transversals. Now weβre ready to look at some examples to see how this plays out.

Using the information in the figure, determine the length of line segment πΈπΉ.

First of all, we can identify line segment πΈπΉ. And then we need to think about what we know based on the figure. In the figure, we have three parallel lines. Line π΄π· is parallel to line πΈπ΅, which is parallel to line πΉπΆ. We can also say that lines π·πΉ and π΄πΆ are transversals of the three parallel lines. Based on this, we know that the parallel lines are going to cut the transversals proportionally. This means that line segment π·πΈ over line segment π΄π΅ will be equal to line segment πΈπΉ over line segment π΅πΆ because of the parallel lines and transversal properties. Once we have this statement, we can just plug in the values for the three line segments we know and use that information to solve for the fourth line segment, which will look like this. 48 over 47 is equal to πΈπΉ over 141.

From there, we cross multiply. 141 times 48 must be equal to 47 times πΈπΉ. 6768 is equal to 47 πΈπΉ. And then we divide both sides of this equation by 47, which tells us that 144 is equal to πΈπΉ. And so we can say that line segment πΈπΉ must measure 144 centimeters.

In our next example, weβll be dealing with two transversals, but this time theyβre cut by four parallel lines.

In the figure, lines πΏ one, πΏ two, πΏ three, and πΏ four are all parallel. Given that ππ equals 12, ππ equals eight, π΄π΅ equals 10, and π΅πΆ equals five, what is the length of line segment πΆπ·?

The first thing we can do here is take the information given in our question and label the figure. First of all, we know that lines πΏ one through πΏ four are all parallel. The line segment ππ measures 12, the line segment ππ equals eight, the line segment π΄π΅ equals 10, and the line segment π΅πΆ equals five. πΆπ· is the line segment weβre interested in finding the length of. But thereβs one more thing we should say about the figure. And that is that π prime and π are lines that are transversals. Theyβre lines that cross all of the parallel lines. Because weβre dealing with four parallel lines and then we have transversals, we know that the segments of the transversals will be proportional.

Using the parallel lines and transversal properties, we can say that line segment ππ over line segment πΆπ· will be equal to the length of line segment ππ. And this is where we need to be careful. Weβre setting up a proportion for π prime between πΏ one and πΏ three. And that means the corresponding values from the transversal line π must also be from line one to line three. The corresponding proportional segment would then be from π΄ to πΆ. And thatβs fine because we can add the distance from π΄ to π΅ and the distance from π΅ to πΆ to find the distance from π΄ to πΆ, which is 15.

When we plug in what we know, weβre saying that eight over line segment πΆπ· will be equal to 12 over 15. To solve for πΆπ·, we cross multiply. Eight times 15 is equal to 12 times πΆπ·. 120 equals 12 times πΆπ·. And if we divide both sides by 12, we see that πΆπ· must equal 10. Looking back on our figure, we can see something interesting here, and that is this segment π΄π· has the same length as the segment πΆπ·. And we know that if parallel lines cut congruent segments of one transversal, then they will cut congruent segments for every transversal. And that means we could say that the segment between πΏ three and πΏ four of π prime will be equal to the segment between πΏ one and πΏ two.

We actually could find out that ππ equals eight. And therefore, ππ would equal four. Additionally, we can show that the distance between line one and line two is two times the distance between line two and line three, which is also true of line three and line four. But back to the question in hand, line segment πΆπ· had a measure of 10.

In our next example, weβll look at parallel lines in a polygon, specifically in a triangle.

If πΆπΈ equals π₯ plus two centimeters, what is π₯?

First of all, on our figure, we can label πΆπΈ as π₯ plus two centimeters. And then we should think about what else we know based on the figure. First of all, we see that line segment πΈπ· is parallel to line segment πΆπ΅. And then we can say that line segment π΄π΅ and line segment π΄πΆ are transversals of these two parallel lines. Based on these two facts, we can draw some conclusions. We can say that the parallel lines πΈπ· and πΆπ΅ cut this triangle proportionally. So we can say line segment π΄πΈ over line segment π΄π· will be equal to line segment πΆπΈ over line segment π·π΅ by parallel lines and transversal properties.

To solve then, we can just plug in the values that we know for these line segments. Six over π₯ plus two is equal to four over eight. The first way we could solve this is by using cross multiplication. We can say six times eight is equal to four times π₯ plus two. Therefore, 48 equals four times π₯ plus two. And if we divide both sides of the equation by four, we see that 12 is equal to π₯ plus two. So we subtract two from both sides, and we see that π₯ equals 10. Now, I said this is one way to solve. And thatβs because if we think about proportionality, and we know that the parallel lines cut these line segments proportionally, we notice that line segment π·π΅ is two times line segment π΄π·.

And in order for things to be proportional, that would mean that the same thing would have to be true on the other side. This means that π₯ plus two must be equal to six times two, which again shows us that side length πΆπΈ must be equal to 12 and therefore π₯ plus two must be equal to 12. So again, π₯ equals 10.

In our final example, we have three parallel lines cut by two transversals, but we have two variables to solve for.

In the given figure, find the values of π₯ and π¦.

The first thing we wanna do is identify what the figure tells us. First off, we see that line π½π is parallel to line πΎπ, which is parallel to line πΏπ. And then we can say that the line π½πΏ and the line ππ are transversals of those three parallel lines. We also notice that the length of line segment ππ is equal to ππ. From this information, we can draw a few conclusions. First, because these two transversals are cut by parallel lines, the segments created on the transversals will be proportional. And secondly, since we know that on one transversal the segments are congruent, we can say that the segments on the other transversal will also be congruent.

We can say that π½πΎ must also be equal to πΎπΏ because theyβre segments that are cut by parallel lines. And this means we can set up two equations. We can set up one equation for π½πΎ equals πΎπΏ and another equation for ππ equals ππ. We can say six π₯ minus 20 equals four π₯ minus eight and five π¦ minus 25 equals three π¦ minus seven. In our first equation, we subtract four π₯ from both sides, and we get two π₯ minus 20 equals negative eight. We add 20 to both sides. Two π₯ then equals 12. And we divide both sides by two, to see that π₯ is equal to six.

We can plug this value back into our two expressions. Six π₯ minus 20 equals 16, and four π₯ minus eight also equals 16. Following the same procedure to solve for π¦, we subtract three π¦ from both sides of our equation. And two π¦ minus 25 equals negative seven. We add 25 to both sides, and two π¦ equals 18. Dividing both sides by two, we see that π¦ equals nine. If we plug those values back into our expressions, five π¦ minus 25 equals 20, and three π¦ minus seven equals 20, which means we can say the values that make these expressions true are π₯ equals six and π¦ equals nine.

Before we finish, letβs take a look at the key points from this video. The first property of parallel lines we saw is that if three or more parallel lines intersect two transversals, then they cut the transversals proportionally. This property is also true when dealing with parallel lines and transversals in polygons. And finally, congruent segments on transversals: if three or more parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal.