Lesson Explainer: Parallel Lines and Transversals: Proportional Parts Mathematics

In this explainer, 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.

Definition: Transversal

A transversal is a line that intersects two or more lines in the same plane at distinct points.

The lines that a transversal intersects do not have to be parallel, but in each of the problems we will look at, they are.

Two examples of transversals are 𝐴𝐶 and 𝐷𝐹 in the following figure, because they both intersect each of the three parallel lines at distinct points. We can see that the transversal 𝐴𝐶 intersects the lines at points 𝐴, 𝐵, and 𝐶, while the transversal 𝐷𝐹 intersects them at points 𝐷, 𝐸, and 𝐹.

Notice that the intersections of the three parallel lines and the two transversals create four different line segments. A segment with endpoints at 𝐴 and 𝐵𝐴𝐵 and another segment with endpoints at 𝐵 and 𝐶𝐵𝐶 both lie on the transversal 𝐴𝐶. Likewise, a segment with endpoints at 𝐷 and 𝐸𝐷𝐸 and another segment with endpoints at 𝐸 and 𝐹𝐸𝐹 both lie on the transversal 𝐷𝐹.

When a transversal is cut by parallel lines, the corresponding angles are congruent. Thus, we know that quadrilaterals 𝐴𝐵𝐸𝐷 and 𝐵𝐶𝐹𝐸 in the figure are similar. Each of our four segments is a side of one of these quadrilaterals, with 𝐴𝐵 corresponding to 𝐵𝐶 and 𝐷𝐸 corresponding to 𝐸𝐹. The fact that corresponding sides of similar figures are proportional leads us to a theorem of parallel lines and transversals.

Theorem: The Basic Proportionality Theorem (Thales’s Theorem)

If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.

Based on this theorem, we know that, in our figure, the ratio of the length of 𝐴𝐵 to that of 𝐵𝐶 is equal to the ratio of the length of 𝐷𝐸 to that of 𝐸𝐹. We can write this as the proportion 𝐴𝐵𝐵𝐶=𝐷𝐸𝐸𝐹.

Notice that, in this proportion, the ratio on each side of the equation contains the lengths of segments on the same transversal. It is worth noting that we could also write a proportion in which the ratio on each side of the equation contains the lengths of corresponding segments on the two different transversals. That is, 𝐴𝐵𝐷𝐸=𝐵𝐶𝐸𝐹.

Solving each of these proportions for an unknown segment length will give the same result, because cross-multiplying will lead to either the equation (𝐸𝐹)(𝐴𝐵)=(𝐵𝐶)(𝐷𝐸) or an equivalent equation, regardless of which of the two proportions is used.

Suppose now that our figure had appeared as follows, indicating that 𝐷𝐸 and 𝐸𝐹 are congruent.

This would tell us that 𝐷𝐸=𝐸𝐹, which would allow us to substitute 𝐷𝐸 into the proportion 𝐴𝐵𝐵𝐶=𝐷𝐸𝐸𝐹 for 𝐸𝐹, giving us 𝐴𝐵𝐵𝐶=𝐷𝐸𝐷𝐸.

We would then be able to simplify the right side of the equation to 1, so we would know that 𝐴𝐵=𝐵𝐶. This leads us to another theorem of parallel lines and transversals.

Theorem: Thales’s Special Theorem

If three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.

Let’s now use these theorems in the problems that follow to find missing segment lengths on transversals when two transversals are intersected by three or more lines.

Example 1: Using the Properties of Parallel Lines and Transversals to Find a Missing Side Length

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

Answer

In the figure, we can see that we have three parallel lines: 𝐴𝐷𝐵𝐸𝐶𝐹.

We can also see that these parallel lines are cut by the two transversals 𝐷𝐹 and 𝐴𝐶. Remember that a transversal is a line that intersects two or more lines in the same plane at distinct points. The basic proportionality theorem (Thales’s theorem) tells us that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. For this reason, we know that the ratio of the length of 𝐷𝐸 to that of 𝐸𝐹 must be equivalent to the ratio of the length of 𝐴𝐵 to that of 𝐵𝐶. Thus, we can write the proportion 𝐷𝐸𝐸𝐹=𝐴𝐵𝐵𝐶.

The figure shows us that 𝐷𝐸=48cm, 𝐴𝐵=47cm, and 𝐵𝐶=141cm, so we can substitute these values into the proportion to get 48𝐸𝐹=47141.

Cross-multiplying then gives us the equation (141)(48)=(𝐸𝐹)(47), which we can simplify to 6768=47𝐸𝐹.

Finally, dividing both sides of the equation by 47, we get 144=𝐸𝐹.

Thus, we know that the length of 𝐸𝐹 in the figure is 144 cm.

Note

Another proportion we could write to solve the problem is 𝐷𝐸𝐴𝐵=𝐸𝐹𝐵𝐶.

We can see that, in this proportion, the ratio on each side of the equation contains the lengths of corresponding segments on the two different transversals rather than the lengths of segments on the same transversal. Substituting into this proportion gives us 4847=𝐸𝐹141, and by multiplying both sides of the equation by 141, we arrive at 144=𝐸𝐹.

Thus, we get the same length for 𝐸𝐹 regardless of which of the two proportions we use. We again find that the length of 𝐸𝐹 in the figure is 144 cm.

In the next example, we will also use the basic proportionality theorem (Thales’s theorem) to find the length of a line segment. This time, however, we will also need to use the fact that the length of a line segment is equal to the sum of the lengths of the shorter disjoint segments that form it to solve the problem.

Example 2: Using the Properties of Parallel Lines and Multiple Transversals to Find a Missing Side Length

In the figure, lines 𝐿, 𝐿, 𝐿, and 𝐿 are all parallel. Given that 𝑋𝑍=12, 𝑍𝑁=8, 𝐴𝐵=10, and 𝐵𝐶=5, what is the length of 𝐶𝐷?

Answer

In this problem, we are told that 𝐿𝐿𝐿𝐿.

The lengths of 𝑋𝑍, 𝑍𝑁, 𝐴𝐵, and 𝐵𝐶 are also given to us, so we can label the figure as shown:

We can see that each of the lines 𝐿, 𝐿, 𝐿, and 𝐿 is cut by the two transversals 𝑀 and 𝑀. Recall that the basic proportionality theorem (Thales’s theorem) states that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. This tells us that the ratio of the length of 𝑋𝑍 to that of 𝑍𝑁 must be equal to the ratio of the length of 𝐴𝐶 to that of 𝐶𝐷. This allows us to write the proportion 𝑋𝑍𝑍𝑁=𝐴𝐶𝐶𝐷.

Notice, however, that we are not given the length of either 𝐴𝐶 or 𝐶𝐷. We are given the lengths of both 𝐴𝐵 and 𝐵𝐶, though, so we can use the fact that the length of a line segment is equal to the sum of the lengths of the shorter disjoint segments that form it to rewrite the proportion in a way that will allow us to solve the problem. Since 𝐴𝐶=𝐴𝐵+𝐵𝐶, we get 𝑋𝑍𝑍𝑁=𝐴𝐵+𝐵𝐶𝐶𝐷.

After substituting the given values of 𝑋𝑍, 𝑍𝑁, 𝐴𝐵, and 𝐵𝐶 into this proportion, we then arrive at 128=10+5𝐶𝐷. which we can rewrite as 128=15𝐶𝐷.

Next, cross-multiplying gives us the equation (𝐶𝐷)(12)=(8)(15). which we can simplify to 12𝐶𝐷=120.

Finally, to solve for 𝐶𝐷, we can divide both sides of the equation by 12 to get 𝐶𝐷=10.

We can therefore conclude that the length of 𝐶𝐷 in the figure is 10.

Next, let’s look at a problem in which we can use Thales’s special theorem to determine a line segment’s length.

Example 3: Finding the Lengths of Proportional Line Segments using the Properties of Parallel Lines and Multiple Traversals

Given that 𝑋𝐿=9cm, find the length of 𝑋𝑍.

Answer

We can see that, in this problem, we have four parallel lines: 𝐴𝑋𝐵𝑌𝐶𝑍𝐷𝐿.

We can also see that these parallel lines cut off 𝐴𝐵, 𝐵𝐶, and 𝐶𝐷 on the transversal 𝐴𝐷, with 𝐴𝐵𝐵𝐶𝐶𝐷, as well as 𝑋𝑌, 𝑌𝑍, and 𝑍𝐿 on the transversal 𝑋𝐿, with the length of 𝑋𝐿 being 9 cm. Based on this information, we are asked to find the length of 𝑋𝑍.

Remember that Thales’s special theorem tells us that if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. Thus, since we know that 𝐴𝐵𝐵𝐶𝐶𝐷, we also know that 𝑋𝑌𝑌𝑍𝑍𝐿.

To begin, we can use the fact that the length of a line segment is equal to the sum of the lengths of the shorter disjoint segments that form it to write the equation 𝑋𝐿=𝑋𝑌+𝑌𝑍+𝑍𝐿.

Since 𝑋𝑌𝑌𝑍𝑍𝐿, we know that 𝑋𝑌=𝑌𝑍=𝑍𝐿. This allows us to substitute 𝑋𝑌 into the equation 𝑋𝐿=𝑋𝑌+𝑌𝑍+𝑍𝐿 for both 𝑌𝑍 and 𝑍𝐿 to get 𝑋𝐿=𝑋𝑌+𝑋𝑌+𝑋𝑌, which we can then rewrite as 𝑋𝐿=3𝑋𝑌.

Recall that we were told that the length of 𝑋𝐿 is 9 cm, so we can now substitute 9 into the equation 𝑋𝐿=3𝑋𝑌 for 𝑋𝐿 to get 9=3𝑋𝑌 and divide both sides by 3, giving us 𝑋𝑌=3.

Remember that 𝑋𝑌 and 𝑌𝑍 have the same length, so we also know that 𝑌𝑍=3.

Next, because the length of a line segment is equal to the sum of the lengths of the segments of which it is comprised, we can write the equation 𝑋𝑍=𝑋𝑌+𝑌𝑍, and after substituting 3 for both 𝑋𝑌 and 𝑌𝑍, we arrive at 𝑋𝑍=3+3,𝑋𝑍=6.or

Thus, the length of 𝑋𝑍 in the figure is 6 cm.

Note

An alternate solution method is to use the basic proportionality theorem (Thales’s theorem). According to this theorem, if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. In order to find the length of 𝑋𝑍 using this theorem, we must set up and solve a proportion using the information we have been given.

Let’s begin by assuming that the length of 𝐴𝐷 is 𝑛. That is, 𝐴𝐷=𝑛.

Because the length of a line segment is equal to the sum of the lengths of the shorter disjoint segments that form it, we can write the equation 𝐴𝐷=𝐴𝐵+𝐵𝐶+𝐶𝐷.

Substituting 𝐴𝐵+𝐵𝐶+𝐶𝐷 into the equation 𝐴𝐷=𝑛 for 𝐴𝐷 then gives us 𝐴𝐵+𝐵𝐶+𝐶𝐷=𝑛.

Since 𝐴𝐵𝐵𝐶𝐶𝐷, we know that 𝐴𝐵=𝐵𝐶=𝐶𝐷. Therefore, we can now substitute 𝐴𝐵 into the equation 𝐴𝐵+𝐵𝐶+𝐶𝐷=𝑛 for both 𝐵𝐶 and 𝐶𝐷 to get 𝐴𝐵+𝐴𝐵+𝐴𝐵=𝑛, which we can rewrite as 3𝐴𝐵=𝑛.

Next, dividing both sides by 3, we arrive at 𝐴𝐵=13𝑛.

Since we were told that 𝐴𝐵 is congruent to 𝐵𝐶, we also know that 𝐵𝐶=13𝑛. Thus, again using the fact that the length of a line segment is equal to the sum of the lengths of the segments of which it is comprised, we can determine the length of𝐴𝐶 in terms of 𝑛: 𝐴𝐶=𝐴𝐵+𝐵𝐶=13𝑛+13𝑛=23𝑛.

Now that we know that the length of 𝐴𝐷 is 𝑛 and that the length of 𝐴𝐶 is 23𝑛, we can set up a proportion that we can then solve to find the length of 𝑋𝑍. In the figure, we can see that 𝐴𝐶 corresponds to 𝑋𝑍 and that 𝐴𝐷 corresponds to 𝑋𝐿, so we can write 𝐴𝐶𝐴𝐷=𝑋𝑍𝑋𝐿.

Next, substituting 23𝑛 for 𝐴𝐶, 𝑛 for 𝐴𝐷, and 9 for 𝑋𝐿, we get 𝑛𝑛=𝑋𝑍9.

This equation can be simplified to 23=𝑋𝑍9, and after multiplying both sides by 9, we arrive at 6=𝑋𝑍.

Thus, we have again determined that the length of 𝑋𝑍 in the figure is 6 cm.

The lengths of line segments are not always integers. They can also be expressed in terms of variables. In the next problem, we will look at an example of this, and we will solve for the variable, using the basic proportionality theorem (Thales’s theorem) as we do so.

Example 4: Using the Properties of Parallel Lines and Solving Linear Equations to Determine the Value of an Unknown and the Length of a Line Segment

In the diagram below, 𝐴𝐵=10, 𝐵𝐶=(𝑥+1), 𝐶𝐷=20, 𝐸𝐹=10, and 𝐹𝐺=10. Find the value of 𝑥 and the length of 𝐺𝐻.

Answer

Let’s begin by using the lengths of 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, 𝐸𝐹, and 𝐹𝐺 that have been given to us to label the diagram as shown:

We can see that, in the diagram, we have four parallel lines: 𝐴𝐸𝐵𝐹𝐶𝐺𝐷𝐻.

We can also see that each of these lines is cut by the two transversals 𝐴𝐷 and 𝐸𝐻. The basic proportionality theorem (Thales’s theorem) tells us that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. Thus, we know that that the ratio of the length of 𝐴𝐵 to that of 𝐵𝐶 must be equal to the ratio of the length of 𝐸𝐹 to that of 𝐹𝐺. This allows us to write the proportion 𝐴𝐵𝐵𝐶=𝐸𝐹𝐹𝐺.

Substituting the given lengths for 𝐴𝐵, 𝐵𝐶, 𝐸𝐹, and 𝐹𝐺 into this proportion then gives us 10𝑥+1=1010, and after simplifying the right side, we get 10𝑥+1=1.

Now, to eliminate the denominator on the left side, we can multiply both sides by 𝑥+1 to get 10=𝑥+1, and to solve the resulting equation for 𝑥, we can subtract 1 from both sides to get 𝑥=9.

Next, to find the length of 𝐺𝐻, we can again use the basic proportionality theorem (Thales’s theorem) to write the proportion 𝐴𝐵𝐶𝐷=𝐸𝐹𝐺𝐻.

Substituting the given values of 𝐴𝐵, 𝐶𝐷, and 𝐸𝐹 into this proportion then gives us 1020=10𝐺𝐻.

Notice that the numerators of the fractions on both sides of the proportion are the same. We, therefore, know that the denominators of the fractions must also be the same, so 𝐺𝐻=20.

In summary, in the diagram, we can deduce that 𝑥=9 and 𝐺𝐻=20.

As a final example, we will solve for two variables instead of one, this time using Thales’s special theorem to help us find the answer.

Example 5: Finding the Lengths of Sides Using the Properties of Parallel Lines to Form Linear Equations

In the given figure, find the values of 𝑥 and 𝑦.

Answer

By examining the figure, we can see that we have three parallel lines: 𝐽𝑀𝐾𝑃𝐿𝑄.

We can also see that these parallel lines cut off 𝐽𝐾 and 𝐾𝐿 on the transversal 𝐽𝐿, as well as 𝑀𝑃 and 𝑃𝑄 on the transversal 𝑀𝑄, with 𝑀𝑃𝑃𝑄.

Remember that the Thales’s special theorem tells us that if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal. Thus, since we know that 𝑀𝑃𝑃𝑄, we also know that 𝐽𝐾𝐾𝐿.

Congruent segments have lengths that are equal, so it follows that 𝑀𝑃=𝑃𝑄𝐽𝐾=𝐾𝐿.andthat

The figure shows us that 𝐽𝐾=6𝑥20, 𝐾𝐿=4𝑥8, 𝑀𝑃=5𝑦25, and 𝑃𝑄=3𝑦7, so we can substitute the appropriate expression from the figure for each segment length, giving us the equations 5𝑦25=3𝑦76𝑥20=4𝑥8.and

Let’s solve each of these equations in turn. First, to solve the equation 6𝑥20=4𝑥8, we can subtract 4𝑥 from both sides to get 2𝑥20=8, and then add 20 to both sides so that our equation becomes 2𝑥=12.

Finally, we can divide both sides by 2 to arrive at 𝑥=6.

Next, to solve the equation 5𝑦25=3𝑦7, we can subtract 3𝑦 from both sides to get 2𝑦25=7, and then add 25 to both sides so that our equation becomes 2𝑦=18.

Finally, we can divide both sides by 2 to arrive at 𝑦=9.

Thus, in the figure we can deduce that 𝑥=6 and 𝑦=9.

Note

An alternate solution method is to use the basic proportionality theorem (Thales’s theorem). Recall that this theorem tells us that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. For this reason, we know that the ratio of the length of 𝐽𝐾 to that of 𝐾𝐿 must be equivalent to the ratio of the length of 𝑀𝑃 to that of 𝑃𝑄. Therefore, we can write the proportion 𝐽𝐾𝐾𝐿=𝑀𝑃𝑃𝑄.

We can then substitute the appropriate expression into the proportion for each segment’s length to get 6𝑥204𝑥8=5𝑦253𝑦7.

Because we know that 𝑀𝑃=𝑃𝑄, we also know that 𝑀𝑃𝑃𝑄=1. It follows that 6𝑥204𝑥8=5𝑦253𝑦7=1, which we can rewrite as the separate equations 6𝑥204𝑥8=15𝑦253𝑦7=1.and

To solve the equation 6𝑥204𝑥8=1, we can begin by multiplying both sides by 4𝑥8 to eliminate the denominator on the left side. This gives us 6𝑥20=4𝑥8, which is the same equation we solved with our previous method to find the value of 𝑥.

To solve the equation 5𝑦253𝑦7=1, we can again begin by eliminating the denominator, this time by multiplying both sides of the equation by 3𝑦7. After doing so, we get 5𝑦25=3𝑦7, which is the same equation we solved with our previous method to find the value of 𝑦. Therefore, we will get the same 𝑥- and 𝑦-values by using this method. Again, we will find that 𝑥=6 and 𝑦=9.

Now, let’s finish by recapping some key points.

Key Points

  • A transversal is a line that intersects two or more lines in the same plane at distinct points.
  • The basic proportionality theorem (Thales’s theorem) states that if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.
  • Thales’s special theorem states that if three or more parallel lines cut off congruent segments on one transversal, then they cut off congruent segments on every transversal.
  • The length of a line segment is equal to the sum of the lengths of the shorter disjoint segments that form it.

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