Lesson Explainer: Parallel Lines in a Triangle | Nagwa Lesson Explainer: Parallel Lines in a Triangle | Nagwa

ู†ู‚ุงุท ุงู„ู…ูƒุงูุขุช

ุงูƒุณุจ ู†ู‚ุงุทู‹ุง ู…ู† ุฎู„ุงู„ ุชูุงุนู„ูƒ ู…ุน ุงู„ู…ูุฏุฑุณ ุฃุซู†ุงุก ุงู„ุญุตุฉ ูˆุฅุฌุงุจุชูƒ ุนู† ุงู„ุฃุณุฆู„ุฉุŒ ูˆุงุณุชุฎุฏู…ู‡ุง ู„ุงุญู‚ู‹ุง ู„ู„ุญุตูˆู„ ุนู„ู‰ ุฎุตู… ุนู†ุฏ ุดุฑุงุก ุญุตุต ุฃุฎุฑู‰.

Lesson Explainer: Parallel Lines in a Triangle Mathematics • First Year of Secondary School

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In this explainer, we will learn how to find missing lengths in a triangle containing two or three parallel lines using proportionality.

Recall that when two parallel lines are cut by a transversal, the resulting corresponding angles are equal.

By adding a second transversal as pictured below, we can form two triangles.

Giving each vertex a label, we can define the larger triangle โ–ณ๐ด๐ท๐ธ and the smaller triangle โ–ณ๐ด๐ต๐ถ.

Since the two pairs of corresponding angles are equal, triangle ๐ด๐ท๐ธ is similar to triangle ๐ด๐ต๐ถ: โ–ณ๐ด๐ท๐ธโˆผโ–ณ๐ด๐ต๐ถ.

Since these triangles are similar, the ratios of their corresponding side lengths must be equal. In other words, we have ๐ด๐ต๐ด๐ท=๐ด๐ถ๐ด๐ธ=๐ต๐ถ๐ท๐ธ.

In the first example, we will demonstrate how to use this definition of the similarity of triangles to identify which pairs of side lengths have equal proportions when a triangle is cut by a line parallel to one of its sides.

Example 1: Identifying Proportions in Triangles

Using the diagram, which of the following is equal to ๐ด๐ต๐ด๐ท?

  1. ๐ด๐ถ๐ธ๐ถ
  2. ๐ด๐ต๐ท๐ต
  3. ๐ด๐ท๐ท๐ต
  4. ๐ด๐ถ๐ด๐ธ
  5. ๐ด๐ธ๐ธ๐ถ

Answer

The diagram indicates that ๐ธ๐ท is parallel to ๐ถ๐ต. Since corresponding angles are equal, that is, โˆ ๐ท๐ธ๐ด=โˆ ๐ต๐ถ๐ด and โˆ ๐ธ๐ท๐ด=โˆ ๐ถ๐ต๐ด,๐ธ๐ท creates triangle ๐ด๐ท๐ธ that is similar to the larger triangle ๐ด๐ต๐ถ.

Since these triangles are similar, the ratios of their corresponding side lengths must be equal. In particular, ๐ด๐ธ๐ด๐ถ=๐ด๐ท๐ด๐ต.

To find the fraction that is equivalent to ๐ด๐ต๐ด๐ท, we can find the reciprocal of both sides of this equation: ๐ด๐ถ๐ด๐ธ=๐ด๐ต๐ด๐ท.

๐ด๐ถ๐ด๐ธ is equal to ๐ด๐ต๐ด๐ท.

Example 2: Finding an Unknown Length in a Triangle Using Proportions

Find the value of ๐‘ฅ.

Answer

๏ƒซ๐ด๐ถ and ๏ƒซ๐ด๐ต are transversals that intersect parallel lines โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ท๐ธ and โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ต๐ถ. Since the two pairs of corresponding angles created by this intersection are equal, that is, โˆ ๐ท๐ธ๐ด=โˆ ๐ต๐ถ๐ด,โˆ ๐ธ๐ท๐ด=โˆ ๐ถ๐ต๐ด, we can say that triangle ๐ด๐ท๐ธ is similar to triangle ๐ด๐ต๐ถ: โ–ณ๐ด๐ต๐ถโˆผโ–ณ๐ด๐ท๐ธ.

When two triangles are similar, the ratios of the lengths of their corresponding sides are equal. In particular, ๐ด๐ท๐ด๐ต=๐ท๐ธ๐ต๐ถ.

By substituting in known values for the lengths of ๐ด๐ท, ๐ท๐ธ, and ๐ด๐ต (where we should note that ๐ด๐ต is the sum of ๐ด๐ท and ๐ท๐ต), we can find the value of ๐‘ฅ: 1010+11=10๐‘ฅ. Solving for ๐‘ฅ, ๐‘ฅ=21.

In the previous two examples, we noted that, if a line intersecting two sides of a triangle is parallel to the third side, then the smaller triangle created by the parallel line is similar to the original triangle. We recall the diagram we presented earlier.

Since triangles ๐ด๐ต๐ถ and ๐ด๐ท๐ธ are similar, we obtain the equal proportions: ๐ด๐ต๐ด๐ท=๐ด๐ถ๐ด๐ธ.

From this diagram, we also note that the line segments ๐ด๐ท and ๐ด๐ธ can be split as follows: ๐ด๐ท=๐ด๐ต+๐ต๐ท๐ด๐ธ=๐ด๐ถ+๐ถ๐ธ.and

Substituting these expressions into our earlier equation and rearranging, ๐ด๐ต๐ด๐ท=๐ด๐ถ๐ด๐ธ๐ด๐ต๐ด๐ต+๐ต๐ท=๐ด๐ถ๐ด๐ถ+๐ถ๐ธ๐ด๐ต(๐ด๐ถ+๐ถ๐ธ)=๐ด๐ถ(๐ด๐ต+๐ต๐ท)๐ด๐ตโ‹…๐ด๐ถ+๐ด๐ตโ‹…๐ถ๐ธ=๐ด๐ถโ‹…๐ด๐ต+๐ด๐ถโ‹…๐ต๐ท.

We can now subtract ๐ด๐ตโ‹…๐ด๐ถ from both sides to find ๐ด๐ตโ‹…๐ถ๐ธ=๐ด๐ถโ‹…๐ต๐ท,๐ด๐ต๐ต๐ท=๐ด๐ถ๐ถ๐ธ.

This leads us to the definition of a theorem that links the line segments created when a parallel side is added to a triangle.

Theorem: Side Splitter Theorem

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally.

Note:

The side splitter theorem can be extended to include parallel lines that lie outside of the triangle. When a straight line lies outside of a triangle and is parallel to one side of the triangle, it forms another triangle that is similar to the first one. This is demonstrated in the following diagram. In this case, an analog of the side splitter theorem can be deduced directly from the similar triangles.

In our next example, we will see how to use this theorem to identify proportional segments of triangles to calculate a missing length.

Example 3: Using Proportions in a Triangle to Calculate an Unknown Length

In the figure, ๐‘‹๐‘Œ and ๐ต๐ถ are parallel. If ๐ด๐‘‹=18, ๐‘‹๐ต=24, and ๐ด๐‘Œ=27, what is the length of ๐‘Œ๐ถ?

Answer

We are given that ๐‘‹๐‘Œ is parallel to ๐ต๐ถ. The side splitter theorem says that if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally.

In particular, ๐ด๐‘Œ๐‘Œ๐ถ=๐ด๐‘‹๐‘‹๐ต.

Substituting ๐ด๐‘‹=18, ๐‘‹๐ต=24, and ๐ด๐‘Œ=27 into this equation and solving for ๐‘Œ๐ถ, 27๐‘Œ๐ถ=1824๐‘Œ๐ถ27=2418๐‘Œ๐ถ=2418ร—27=36.

The length of ๐‘Œ๐ถ is 36.

In our next example, we will demonstrate how to solve multistep problems involving triangles and parallel lines.

Example 4: Finding Unknowns in an Applied Problem

The given figure shows a triangle ๐ด๐ต๐ถ.

  1. Work out the value of ๐‘ฅ.
  2. Work out the value of ๐‘ฆ.

Answer

Part 1

In the figure, a line parallel to side ๐ต๐ถ is intersecting the other two sides of the triangle. The side splitter theorem tells us that this line divides those sides proportionally.

Labelling this line segment as ๐ท๐ธ, we obtain ๐ด๐ท๐ท๐ต=๐ด๐ธ๐ธ๐ถ.

This gives us an equation that can be solved for ๐‘ฅ: 32๐‘ฅ+3=2๐‘ฅ+53(๐‘ฅ+5)=2(2๐‘ฅ+3)3๐‘ฅ+15=4๐‘ฅ+615=๐‘ฅ+6๐‘ฅ=9.

Part 2

Now that we know the value of ๐‘ฅ, we can use this information to find the value of ๐‘ฆ. Since the two pairs of corresponding angles created by the intersection of ๐ท๐ธ are equal, triangle ๐ด๐ต๐ถ is similar to triangle ๐ด๐ท๐ธ: โ–ณ๐ด๐ต๐ถโˆผโ–ณ๐ด๐ท๐ธ.

In particular, ๐ด๐ท๐ด๐ต=๐ท๐ธ๐ต๐ถ.

The length of ๐ด๐ต is the sum of the lengths of ๐ด๐ท and ๐ท๐ต. We are given that ๐ด๐ท=3 and ๐ท๐ต=2๐‘ฅ+3. Since ๐‘ฅ=9, ๐ท๐ต=21. Therefore, ๐ด๐ต=3+21=24.

Substituting these values into our earlier equation and solving for ๐‘ฆ, 324=2๐‘ฆ๐‘ฆ24=23๐‘ฆ=23ร—24=16.

Therefore, ๐‘ฆ=16.

In our next example, we will demonstrate how to apply the side splitter theorem to a triangle which contains several pairs of parallel lines.

Example 5: Finding a Side Length in a Triangle Using the Relation between Parallel Lines

Find the length of ๐ถ๐ต.

Answer

From the given diagram we note that ๐ท๐น is parallel to ๐ด๐ธ in the triangle ๐ถ๐ด๐ธ, and ๐ท๐ธ is parallel to ๐ด๐ต in the triangle ๐ถ๐ด๐ต. The side splitter theorem tells us that if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally. Applying this theorem to triangle ๐ถ๐ด๐ธ where ๐ท๐น is parallel to one side of the triangle, we obtain ๐ถ๐น๐น๐ธ=๐ถ๐ท๐ท๐ด.

Since ๐ท๐ธ is parallel to one side of the larger triangle ๐ถ๐ด๐ต, we can also obtain ๐ถ๐ธ๐ธ๐ต=๐ถ๐ท๐ท๐ด.

Both ๐ถ๐น๐น๐ธ and ๐ถ๐ธ๐ธ๐ต are equal to ๐ถ๐ท๐ท๐ด. This means we can set ๐ถ๐น๐น๐ธ=๐ถ๐ธ๐ธ๐ต.

We can substitute the given values ๐ถ๐น=15, ๐น๐ธ=6, and ๐ถ๐ธ=15+6=21 into this equation to obtain an equation that can be solved for ๐ธ๐ต: 156=21๐ธ๐ต๐ธ๐ต=21ร—615.

Therefore, ๐ธ๐ต=8.4.cm

Since ๐ถ๐ต=๐ถ๐น+๐น๐ธ+๐ธ๐ต, ๐ถ๐ต=15+6+8.4=29.4.cm

The length of ๐ถ๐ต is 29.4 cm.

Recall that the side splitter theorem tells us that if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally. Moreover, we have learned that this theorem can be extended to include parallel lines that lie outside of the triangle. It turns out that the converse of this result is also true, which proves very useful when solving problems of this type.

Theorem: The Converse of the Side Splitter Theorem

If a line intersects two sides of a triangle and splits those sides in equal proportions, then that line must be parallel to the third side of the triangle.

In all three diagrams above, ๐ด๐ต๐ถ is a triangle and โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ท๐ธ intersects โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ด๐ต at ๐ท and โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ด๐ถ at ๐ธ.

If ๐ด๐ท๐ท๐ต=๐ด๐ธ๐ธ๐ถ, then โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ท๐ธ must be parallel to โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐ต๐ถ.

By applying the converse of the side splitter theorem, we are able to prove that a straight line is parallel to one side of a triangle due to having proportional parts. In our final example, we will demonstrate this process.

Example 6: Finding the Unknown Lengths in a Triangle given the Other Sidesโ€™ Lengths Using the Relations of Parallel Lines

Given that ๐ด๐ต๐ถ๐ท is a parallelogram, find the length of ๐‘Œ๐‘.

Answer

To find the length of ๐‘Œ๐‘, we will begin by identifying relevant information about triangles ๐‘‹๐‘Œ๐‘ and ๐‘‹๐ท๐ถ. We are given that ๐‘‹๐‘Œ=๐‘Œ๐ท and ๐‘‹๐‘=๐‘๐ถ. We also recall that the side splitter theorem tells us that if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally. Conversely, if a line splits two sides of a triangle into equal proportions, then that line must be parallel to the third side. Since sides ๐‘‹๐ท and ๐‘‹๐ถ of the larger triangle ๐‘‹๐ท๐ถ have been divided into equal proportions, we can apply the converse of this theorem to deduce that ๐ท๐ถ and ๐‘Œ๐‘ must be parallel.

We also recall that if a line parallel to a side of a triangle intersects two other sides, then the smaller triangle created by the parallel line is similar to the original triangle. Hence, we obtain โ–ณ๐‘‹๐‘Œ๐‘โˆผโ–ณ๐‘‹๐ท๐ถ.

Since ๐ท๐ถ is the opposite side of ๐ด๐ต in the parallelogram ๐ด๐ต๐ถ๐ท, these two sides must have the same lengths. Hence, the length of ๐ท๐ถ is 134.9 cm. Denoting the length of ๐‘‹๐‘Œby an unknown constant ๐‘ฅ, we can draw the following diagram.

Since triangles ๐‘‹๐‘Œ๐‘ and ๐‘‹๐ท๐ถ are similar, we can form an equation that links the lengths of the sides ๐‘‹๐‘Œ, ๐‘‹๐ท, ๐‘Œ๐‘, and ๐ท๐ถ: ๐‘‹๐‘Œ๐‘‹๐ท=๐‘Œ๐‘๐ท๐ถ๐‘ฅ2๐‘ฅ=๐‘Œ๐‘134.912=๐‘Œ๐‘134.9.

Solving for ๐‘Œ๐‘, we find ๐‘Œ๐‘=134.92=67.45.

The length of ๐‘Œ๐‘ is 67.45 cm.

We will now recap the key points from this explainer.

Key Points

  • If a line intersecting two sides of a triangle is parallel to the remaining side, then the smaller triangle created by the parallel line is similar to the larger, original triangle.
  • The side splitter theorem tells us that if a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides those sides proportionally.
  • The side splitter theorem can be extended to include parallel lines that lie outside a triangle. If a line lying outside a triangle is parallel to one side of the triangle and intersects the extensions of the other two sides of the triangle, then the line divides the extensions of those sides proportionally.
  • The converse of the side splitter theorem states that if a line splits two sides of a triangle proportionally, then that line is parallel to the remaining side.

ุงู†ุถู… ุฅู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ

ุดุงุฑูƒ ููŠ ุงู„ุญุตุต ุงู„ู…ุจุงุดุฑุฉ ุนู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ ูˆุญู‚ู‚ ุงู„ุชู…ูŠุฒ ุงู„ุฏุฑุงุณูŠ ุจุฅุฑุดุงุฏ ูˆุชูˆุฌูŠู‡ ู…ู† ู…ุฏุฑุณ ุฎุจูŠุฑ!

  • ุญุตุต ุชูุงุนู„ูŠุฉ
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  • ุฃุณุฆู„ุฉ ุงู…ุชุญุงู†ุงุช ูˆุงู‚ุนูŠุฉ

ุชุณุชุฎุฏู… «ู†ุฌูˆู‰» ู…ู„ูุงุช ุชุนุฑูŠู ุงู„ุงุฑุชุจุงุท ู„ุถู…ุงู† ุญุตูˆู„ูƒ ุนู„ู‰ ุฃูุถู„ ุชุฌุฑุจุฉ ุนู„ู‰ ู…ูˆู‚ุนู†ุง. ุงุนุฑู ุงู„ู…ุฒูŠุฏ ุนู† ุณูŠุงุณุฉ ุงู„ุฎุตูˆุตูŠุฉ