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### 3.2. Proportional Segments

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• If two segments (AC and AD below) are divided proportionally (by drawing BE parallel to CD), then resulting corresponding segments are proportional (their ratios are equal) to each other and to the original segments.

Segments AC and AD are divided proportionally, and the resulting segments, AB and BC, as well as AE and ED, are proportional to each other and to original segments AC and AD, so that:

AC/AB is proportional to AD/AE: AC/AB = AD/AE

AC/BC is proportional to AD/ED: AC/BC = AD/ED

BC/AB is proportional to ED/AE: BC/AB = ED/AE

Corresponding ratios are equal.

• Side note: In the above drawing, two similar triangles ΔADC and ΔAEB are formed. Two similar triangles can be created by drawing a line parallel to one of the sides of a triangle that intersects the other two sides. In the figure, ΔADC is similar to ΔAEB because the three corresponding angles are equal or congruent: ∠CAD ≅ ∠BAE, ∠ACD ≅ ∠ABE, and ∠ADC ≅ ∠AEB. (See Section 4.6. Similar Triangles.)

• In the previous section, we showed that the equation (a)(d) = (b)(c) can be rearranged using algebra by multiplying and dividing both sides of the equation by a, b, c, or d, into equivalent forms of proportions:

These proportions can represent proportionally-divided segments, and they correspond to the following four figures and eight proportion equations:

Notice that these figures representing proportionally-divided segments are triangles with a segment drawn parallel to one side. The following theorems apply to the triangle.

• Triangle proportionality theorem also called the side-splitter theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the other two sides proportionally.

| If DE ‖ BC, then a/b = c/d.
If a/b = c/d, then DE ‖ BC. If DE ‖ BC, then a/AB = c/AC and b/AB = d/AC |

Converse of triangle proportionality theorem: If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

Corollary to triangle proportionality theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it produces segments proportional to the side lengths.

If DE ‖ BC, then a/AB = c/AC and b/AB = d/AC.

• Example: Prove the triangle proportionality theorem, which states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides the other two sides proportionally.

| If DE ‖ BC, then a/b = c/d. |

Given: | ΔABC having DE ‖ BC. |

Prove: | a/b = c/d. |

Plan: | This proof uses the similarity of ΔABC and ΔADE in which ΔADE is created by DE being parallel to BC. In the proof show that ΔADE ~ ΔABC. Then show that AB/a = AC/c and by substitution is equivalent to b/a = d/c, which by the inversion property (or algebra) is equivalent to a/b = c/d. (This proof uses principles of similar triangles discussed in Chapter 4, Section 4.6. Similar Triangles.) |

Proof:

Statements | Reasons | ||
---|---|---|---|

1. | ΔABC with DE ‖ BC. | 1. | Given. |

2. | ∠ADE ≅ ∠ABC; ∠AED ≅ACB. | 2. | If two parallel lines intersected by transversal corresponding ∠s are ≅. |

3. | ΔADE ~ ΔABC. | 3. | AA similarity post: If two ∠s of one Δ are ≅ two ∠s of other Δ, then Δs are similar. |

4. | AB/a = AC/c. | 4. | Corresponding sides of similar triangles arein proportion. |

5. | a + b = AB; c + d = AC. | 5. | Segment addition postulate. |

6. | (a + b)/a = (c + d)/c. | 6. | Substitution of #5 into #4. |

7. | b/a = d/c. | 7. | Addition property of a proportion. (a + b)/a = (c + d)/c => 1 + b/a = 1 + d/c => b/a = d/c. |

8. | b/a = d/c equals a/b = c/d. | 8. | Inversion property of proportion. |

• Example: In the triangle ΔABC, if a = 10, b = 6, c = 12, and DE ‖ BC, find d.

| If DE ‖ BC, then a/b = c/d. |

We can use the triangle proportionality theorem: If DE ‖ BC, then a/b = c/d. Then substitute given values into the proportion for a, b, and c, and solve for d:

10/6 = 12/d

Cross-multiply: 10d = 72

d = 7.2

Therefore, d = 7.2.

• Example: In triangle ΔABC, if DE ‖ BC, FE ‖ DC, AF = 2, and FD = 3, find DB.

We can use the triangle proportionality theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides the other two sides proportionally.

In ΔABC, because DE ‖ BC, then AD/DB = AE/EC, or 5/DB = AE/EC

In ΔADC, because FE ‖ DC, then AF/FD = AE/EC, or 2/3 = AE/EC

Combine the two proportions. Because the right sides of the proportion are equal, AE/EC = AE/EC, then, the left sides are equal: 5/DB = 2/3

Solve for DB by first cross-multiplying:

2DB = (5)(3) = 15

DB = 15/2 = 7.5

Therefore, DB = 7.5.

• We have learned that parallel lines in a triangle divide the sides proportionally. In a similar manner, any three parallel lines with two transversals crossing them divide the transversals proportionally.

| a/b = c/d |

Three parallel lines transversals theorem: Three or more parallel lines divide any two transversals intersecting them proportionally.

In the figure a/b is proportional to c/d: a/b = c/d.

• Example: Prove the three parallel lines transversals theorem.

Given: AB ‖ CD ‖ EF.

Prove: a/b = c/d.

Plan: Create ΔEBF and ΔAEB. Use triangle proportionality theorem to show a/b = e/f and e/f = c/d, therefore by substitution a/b = c/d.

Proof:

Statements | Reasons | ||
---|---|---|---|

1. | AB ‖ CD ‖ EF | 1. | Given |

2. | Draw BE across CD forming ΔEBF & ΔAEB. Label segments e & f. | 2. | Through 2 points exists one line. |

3. | a/b = e/f; e/f = c/d. | 3. | Δ proportionality theorem: If a line is ‖ to one side of a Δ, then it divides the other two sides proportionally. |

4. | a/b = c/d. | 4. | Transitive axiom: Things equal to the same are equal to each other. (#3) |

• Example: In the figure, if b = 3, c = 4, and c + d = 6, what is a?

First find d using c + d = 6 and c = 4:

4 + d = 6

d = 6 – 4 = 2

We can use the three parallel lines transversals theorem and substitute into the proportion a/b = c/d (formed by the three parallel lines cut by two transversals):

a/3 = 4/2

2a = (4)(3)

a = 12/2 = 6

Therefore, a = 6 in the proportion a/b = c/d, or 6/3 = 4/2.

• If parallel lines are equidistant from each other, they will divide a transversal into congruent segments.

• Three parallel lines congruent segments theorem: If three parallel lines create congruent segments on one transversal, then they create congruent segments on every transversal.

| If segment a ≅ segment b, then segment c ≅ segment d. |

• Example: In the above figure, if a = 5, c = 4, and d = 4, what is b?

Using the three parallel lines congruent segments theorem, segment c ≅ segment d, therefore segment a ≅ segment b, and a = b = 5.

• A bisector of an angle in a triangle creates a proportion that can be used to determine side lengths.

In ΔPQR angle bisector QS creates ∠α ≅ ∠α and proportion a/b = c/d.

• Triangle angle bisector theorem: A bisector of an angle of a triangle divides the opposite side (of the triangle) into segments, which are proportional to the adjacent sides.

In other words, if a ray or line bisects an angle of a triangle, then it divides the opposite side of the triangle into segments that are proportional to the other two sides.

In the above figure a/b is proportional to c/d, or a/b = c/d.

• Example: Prove the triangle angle bisector theorem:

Given: ΔABC with bisector BD of ∠ABC.

Prove: AB/BC = AD/DC.

Plan: Create ΔAEC with CE ‖ BD. By triangle proportionality theorem, AB/BE = AD/DC. Show that BC ≅ BE and substitute into proportion AB/BE = AD/DC. Use properties of parallel lines and transversals to show BC ≅ BE in isosceles ΔBCE.

Proof:

Statements | Reasons | ||
---|---|---|---|

1. | Draw line CE ‖ BD. | 1. | Through a point not on a line, a line can bedrawn parallel to that line. |

2. | Extend AB to intersect CE at E forming ΔAEC with CE ‖ BD. | 2. | Through 2 points exists 1 line. Also, two lines intersect at exactly 1 point. |

3. | AB/BE = AD/DC. | 3. | Δ proportionality theorem: If a line is ‖ to one side of a Δ, it divides the other two sides proportionally. |

4. | BD bisects ∠ABC. | 4. | Given. |

5. | ∠ABD ≅ ∠DBC. | 5. | Definition of angle bisector. |

6. | ∠DBC ≅ ∠BCE. | 6. | If two ‖ lines (CE ‖ BD) are cut bytransversal (BC), alternate interior angles are ≅. |

7. | ∠ABD ≅ ∠BCE. | 7. | Substitution #5 and #6. |

8. | ∠ABD ≅ ∠AEC. | 8. | If two ‖ lines (CE ‖ BD) are cut bytransversal (AE), corresponding angles are ≅. |

9. | ∠AEC ≅ ∠BCE. | 9. | Substitution #7 and #8. |

10. | BC ≅ BE forming isosceles ΔBCE. | 10. | If 2 angles of a Δ are ≅, then sides opposite those angles are ≅. |

11. | AB/BC = AD/DC. | 11. | Substitution #3 and #10. |

• Example: In ΔPQR, if a = 120 inches, c = 8 feet, and d = 6 feet, find b.

First we must convert to the same units of measure. Let’s convert a to feet: (120 inches)(1 foot/12 inches) = 10 feet (because inches/inches cancels).

Use the triangle angle bisector theorem: A bisector of an angle of a triangle divides the side opposite into segments proportional to the adjacent sides. In the proportion a/b = c/d, we can substitute given values and use algebra to solve for b:

10/b = 8/6

8b = (10)(6) = 60

b = 60/8 = 7.5

Therefore, b = 7.5.

• Example: In the right triangle ΔABC, if leg AB = 9 miles, leg BC = 12 miles, and BD bisects angle B, find AD and DC. (This example uses the Pythagorean Theorem for right triangles discussed in Chapter 4, Section 4.8.)

Using the triangle angle bisector theorem, which states that a bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides: AB/BC = AD/DC

9/12 = AD/DC

Next, we can use the addition property of a proportion:

If a/b = c/d is a proportion, then (a + b)/b = (c + d)/d. Therefore:

(9 + 12)/12 = (AD + DC)/DC

Notice that (AD + DC) is the hypotenuse of right ΔABC, which can be found using the Pythagorean Theorem a^{2} + b^{2} = c^{2}, or equivalently,

AB

^{2}+ BC^{2}= AC^{2}: 9^{2}+ 12^{2}= AC^{2}= 81 + 144 = 225 = AC^{2}

AC = (AD + DC) = 15

Substitute back into (9 + 12)/12 = (AD + DC)/DC:

21/12 = 15/DC

DC(21) = (12)(15)

DC = 180/21 ≈ 8.57

We can use the segment addition postulate AD + DC = AC to find AD:

AD = AC – DC = 15 – 180/21 = 315/21 – 180/21 = 135/21 ≈ 6.43

Therefore, given only the legs we have found AD = 135/21 ≈ 6.43 and DC = 180/21 ≈ 8.57.

To check we can substitute back into the proportion AB/BC = AD/DC:

9/12 = (135/21) / (180/21) => 9/12 = 135/180

(180)(9) = (12)(135), or 1,620 = 1,620