Lesson Video: Parallel Lines and Transversals: Proportional Parts | Nagwa Lesson Video: Parallel Lines and Transversals: Proportional Parts | Nagwa

Lesson Video: Parallel Lines and Transversals: Proportional Parts Mathematics

In this video, we will learn how to use parallelism of lines to find a missing length of a line segment in a transversal line cut by parallel lines.

13:04

Video Transcript

In this video, we will use properties of parallel lines and transversals to find the missing length of a line segment in a transversal line cut by parallel lines. To do that, let’s see how parallel lines can produce proportionality.

A property of parallel lines is that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. Let’s consider what this might look like. We have three parallel lines with two transversals, both of which cross all three parallel lines. Remember that a transversal is a line that crosses at least two other lines. In this figure, we’ll consider these two transversals 𝑙 and π‘š since they are the transversals that cross all three of the parallel lines. The three parallel lines cut by two transversals in this figure creates four line segments.

Here, we have labeled those line segments π‘Ž, 𝑏, 𝑐, 𝑑. By this property, we can say that the ratio π‘Ž to 𝑏 will be equal to the ratio of 𝑐 to 𝑑. Another way to write this proportion is π‘Ž over 𝑐 will be equal to 𝑏 over 𝑑. Before we move on, it’s also worth noting where we might see and use this property. This property is true and applies inside of polygons. We can modify this figure to show an example of this.

If we have the quadrilateral 𝐴𝐡𝐢𝐷, and it’s cut by the line segment 𝐸𝐹 that is parallel to 𝐴𝐷 and 𝐡𝐢, notice that we have three parallel segments cut by two transversals. This means that the segments created will be proportional. This means that segment 𝐴𝐸 over segment 𝐸𝐡 will be equal to segment 𝐷𝐹 over segment 𝐹𝐢.

As an extension of this, there is one more property we need to consider, 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. Here are three parallel lines. And if we have a transversal line 𝑙 such that the two created segments are congruent to one another, then any transversal that is also cut by these three parallel lines will have congruent segments. On the transversal π‘š, the two segments will be congruent to each other. And on the transversal 𝑛, the two created segments are congruent to one another.

But we need to pay close attention to this property. We’re saying that segment π‘Ž is equal to segment 𝑏 or segment 𝑐 is equal to segment 𝑑. The congruency is on the same transversal line, not between lines. You could not say here that segment π‘Ž was equal to segment 𝑐. Now we’re ready to take these properties and apply them to some examples.

Using the information in the figure, determine the length of line segment 𝐸𝐹.

First, we’ll identify line segment 𝐸𝐹. And then we’ll think about what we already know based on the figure. We see in our figure that we have three parallel lines. And so we can say that 𝐴𝐷 is parallel to 𝐸𝐡, which is parallel to 𝐹𝐢. Then we can say that the lines 𝐷𝐹 and 𝐴𝐢 are transversals of the three parallel lines. Because we have three parallel lines being cut by two transversals, we know that the created segments are cut proportionally, by the properties of parallel lines and transversals.

This means we can say that 𝐷𝐸 over 𝐸𝐹 will be equal to 𝐴𝐡 over 𝐡𝐢. If we plug in the lengths we do know, we have 48 over 𝐸𝐹 is equal to 47 over 141. To solve, we cross multiply. 48 times 141 will be equal to 47 times 𝐸𝐹. Therefore, 6768 is equal to 47 times 𝐸𝐹. And dividing both sides by 47 gives us 144 is equal to 𝐸𝐹. Since our segments are measured in centimeters, we can say that 𝐸𝐹 is equal to 144 centimeters.

In our next example, we’ll look at a case where we have two transversals that are cut by four parallel lines.

In the figure, lines 𝐿 one, 𝐿 two, 𝐿 three, and 𝐿 four are parallel. Given that 𝑋𝑍 equals 12, 𝑍𝑁 equals eight, 𝐴𝐡 equals 10, and 𝐡𝐢 equals five, what is the length of segment 𝐢𝐷?

First, let’s take the information we were given and use it to label our figure. We know that 𝐿 one, 𝐿 two, 𝐿 three, and 𝐿 four are parallel. We can add that to our figure. The line segment 𝑋𝑍 equals 12, and 𝑍𝑁 equals eight. 𝐴𝐡 equals 10, and 𝐡𝐢 equals five. The unknown length that we’re trying to solve for is 𝐢𝐷. We can also note that the lines 𝑀 and 𝑀 prime are transversals of all four parallel lines. Because we have three or more parallel lines that are cut by two transversals, we know that the created segments on these transversals will be proportional. Therefore, by parallel lines in transversal properties, we can set up a proportion to solve for the missing segment 𝐢𝐷.

𝑍𝑁 over 𝐢𝐷 is equal to 𝑋𝑍 over 𝐴𝐢. If we plug in what we know, we’ll have eight over 𝐢𝐷 is equal to 12 over 15. To find the distance from 𝐴 to 𝐢, we need to add 10 and five together, which gives us the 15. To solve, we cross multiply. Eight times 15 will be equal to 12 times 𝐢𝐷. 120 equals 12 times 𝐢𝐷. Dividing both sides by 12, we see that 𝐢𝐷 will be equal to 10.

If we add this back to our figure, we notice something interesting. The segment 𝐴𝐡 is equal to the segment 𝐢𝐷. These are congruent segments. And because we know something about congruent segments and transversals, this means we can also say that the segment π‘‹π‘Œ is going to be congruent to the segment 𝑍𝑁. And therefore π‘‹π‘Œ equals eight and π‘Œπ‘ must be equal to four. However, this question was only asking for the length of the line segment 𝐢𝐷. And we’ve shown that 𝐢𝐷 equals 10 length units.

Let’s consider another example.

Given that 𝑋𝐿 equals nine centimeters, find the length of line segment 𝑋𝑍.

First, let’s think about what we see in this image. First of all, we have four parallel lines: 𝐴𝑋, π΅π‘Œ, 𝐢𝑍, and 𝐷𝐿. We can also see that the lines 𝐴𝐷 and 𝑋𝐿 are transversals of the four parallel lines. We also see that on the line 𝐴𝐷, the segments 𝐴𝐡, 𝐡𝐢, and 𝐢𝐷 are congruent to one another. And when three or more lines cut congruent segments on one transversal, we know it will cut congruent segments on all transversals cut by those parallel lines. This is the congruent segments on transversals property, which means the line segment π‘‹π‘Œ is congruent to the segment π‘Œπ‘, which is congruent to the segment 𝑍𝐿.

We also know that the distance from 𝑋 to 𝐿 equals nine centimeters. And since the segment 𝑋𝐿 is made up of three congruent segments, we divide nine by three to find the length of each of the segments. And then we see that π‘‹π‘Œ, π‘Œπ‘, and 𝑍𝐿 will all measure three centimeters. We’re looking for the length of the segment 𝑋𝑍, which is made up of two segments, π‘‹π‘Œ and π‘Œπ‘, both of which measure three centimeters. And so we can say that segment 𝑋𝑍 is equal to six centimeters.

In our final example, we have three parallel lines cut by two transversals, and we’re solving for two unknown values.

In the given figure, find the values of π‘₯ and 𝑦.

First, we see that we have three parallel lines, line 𝐽𝑀, 𝐾𝑃, and 𝐿𝑄. We can also say that the line 𝐽𝐿 and the line 𝑀𝑄 are transversals of the three parallel lines. Additionally, line segment 𝑀𝑃 and line segment 𝑃𝑄 are congruent to one another. Based on these three things, we can say that we have congruent segments on transversals. Since three parallel lines cut congruent segments on one transversal, they will cut congruent segments on the other transversal.

This means that line segment 𝐽𝐾 will be congruent to line segment 𝐾𝐿. And we can mark these congruent segments on our figure. To find π‘₯ and 𝑦, we can then set up two equations. We can say that 𝑀𝑃 is equal to 𝑃𝑄 and 𝐽𝐾 is equal to 𝐾𝐿. Starting with 𝑀𝑃 equals 𝑃𝑄, we substitute five 𝑦 minus 25 in for 𝑀𝑃 and three 𝑦 minus seven in for 𝑃𝑄. We add 25 to both sides of our equation, which will give us five 𝑦 is equal to three 𝑦 plus 18. So we subtract three 𝑦 from both sides to get two 𝑦 equals 18. And finally, dividing both sides by two gives us 𝑦 equals nine.

We’ll follow a similar procedure to solve for π‘₯. We substitute six π‘₯ minus 20 in for 𝐽𝐾 and four π‘₯ minus eight in for 𝐾𝐿. By adding 20 to both sides, we find six π‘₯ equals four π‘₯ plus 12. From there, we subtract four π‘₯ from both sides, which gives us two π‘₯ equals 12. Dividing both sides by two, we find that π‘₯ equals six. We could take the values we found for π‘₯ and 𝑦 and use them to find the length of each line segment. 𝐽𝐾 would be equal to six times six minus 20, which is 16. And because we know that 𝐾𝐿 is congruent to 𝐽𝐾, it would also be equal to 16. And then 𝑀𝑃 would be equal to five times nine minus 25, which is 20. And as 𝑃𝑄 is congruent to 𝑀𝑃, both of these lengths would be 20. However, here we were only asked what the values of π‘₯ and 𝑦 were. And we can say that π‘₯ equals six and 𝑦 equals nine.

Before we finish, let’s quickly review the key points from this video. If three or more parallel lines intersect two transversals, then they cut the transversals proportionally. This property is also true for parallel lines and transversals in polygons. And finally, if three or more parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal.

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