Triangle Proportionality Theorems Worksheet featuring problems to solve for 'x' using proportional segments and completing proportions based on parallel lines.

Problem Analysis:

The worksheet involves using triangle proportionality theorems to solve for unknown values and complete proportions. We will solve each problem step by step.

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Problem 1:

Given:
- A triangle with a line segment parallel to one of its sides.
- The sides are labeled as follows:
- Top side: \(3x - 5\)
- Bottom side: \(10\)
- Right side: \(28\)
- Parallel segment: \(8\)

Using the Basic Proportionality Theorem (Thales' Theorem), which states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally:
\[
\frac{3x - 5}{10} = \frac{28}{8}
\]

#### Step-by-Step Solution:
1. Simplify the right-hand side:
\[
\frac{28}{8} = \frac{7}{2}
\]
So, the equation becomes:
\[
\frac{3x - 5}{10} = \frac{7}{2}
\]

2. Cross-multiply to solve for \(x\):
\[
2(3x - 5) = 10 \cdot 7
\]
\[
6x - 10 = 70
\]

3. Solve for \(x\):
\[
6x = 70 + 10
\]
\[
6x = 80
\]
\[
x = \frac{80}{6} = \frac{40}{3}
\]

#### Final Answer:
\[
\boxed{\frac{40}{3}}
\]

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Problem 2:

Given:
- A triangle with a line segment parallel to one of its sides.
- The sides are labeled as follows:
- Top side: \(8\)
- Bottom side: \(4\)
- Right side: \(12\)
- Parallel segment: \(x\)

Using the Basic Proportionality Theorem:
\[
\frac{8}{4} = \frac{12}{x}
\]

#### Step-by-Step Solution:
1. Simplify the left-hand side:
\[
\frac{8}{4} = 2
\]
So, the equation becomes:
\[
2 = \frac{12}{x}
\]

2. Solve for \(x\):
\[
2x = 12
\]
\[
x = \frac{12}{2} = 6
\]

#### Final Answer:
\[
\boxed{6}
\]

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Problem 3:

Given:
- A triangle with a line segment parallel to one of its sides.
- The sides are labeled as follows:
- Left side: \(8\)
- Bottom side: \(x\)
- Hypotenuse: \(20\)
- Parallel segment: \(18\)

Using the Basic Proportionality Theorem:
\[
\frac{8}{20} = \frac{x}{18}
\]

#### Step-by-Step Solution:
1. Simplify the left-hand side:
\[
\frac{8}{20} = \frac{2}{5}
\]
So, the equation becomes:
\[
\frac{2}{5} = \frac{x}{18}
\]

2. Cross-multiply to solve for \(x\):
\[
2 \cdot 18 = 5x
\]
\[
36 = 5x
\]

3. Solve for \(x\):
\[
x = \frac{36}{5}
\]

#### Final Answer:
\[
\boxed{\frac{36}{5}}
\]

---

Problem 4:

Given:
- A triangle with a line segment parallel to one of its sides.
- The sides are labeled as follows:
- Height from \(A\) to \(D\): \(13 + 8 = 21\)
- Height from \(A\) to \(B\): \(13\)
- Base \(AE\): \(42\)
- Segment \(AU\): \(x\)

Using the Basic Proportionality Theorem:
\[
\frac{AB}{AD} = \frac{AU}{AE}
\]

#### Step-by-Step Solution:
1. Calculate the ratio of the heights:
\[
\frac{AB}{AD} = \frac{13}{21}
\]

2. Use the proportionality theorem:
\[
\frac{13}{21} = \frac{x}{42}
\]

3. Cross-multiply to solve for \(x\):
\[
13 \cdot 42 = 21x
\]
\[
546 = 21x
\]

4. Solve for \(x\):
\[
x = \frac{546}{21} = 26
\]

#### Final Answer:
\[
\boxed{26}
\]

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Problem 5:

Given:
- A figure with multiple parallel lines intersected by transversals.
- Complete the proportions based on the Basic Proportionality Theorem.

#### Step-by-Step Solution:
1. Proportion 1:
\[
\frac{AB}{BD} = \frac{CE}{?}
\]
By the Basic Proportionality Theorem, since \(AB\) and \(BD\) are segments on one transversal and \(CE\) is a corresponding segment on another transversal, the missing term must be \(EG\):
\[
\frac{AB}{BD} = \frac{CE}{EG}
\]

2. Proportion 2:
\[
\frac{FD}{DB} = \frac{EC}{?}
\]
Similarly, by the Basic Proportionality Theorem, the missing term must be \(CA\):
\[
\frac{FD}{DB} = \frac{EC}{CA}
\]

3. Proportion 3:
\[
\frac{BC}{DE} = \frac{AC}{?}
\]
By the Basic Proportionality Theorem, the missing term must be \(AG\):
\[
\frac{BC}{DE} = \frac{AC}{AG}
\]

4. Proportion 4:
\[
\frac{FG}{BC} = \frac{AG}{?}
\]
By the Basic Proportionality Theorem, the missing term must be \(AB\):
\[
\frac{FG}{BC} = \frac{AG}{AB}
\]

#### Final Answers:
\[
\boxed{\text{1) } EG, \text{ 2) } CA, \text{ 3) } AG, \text{ 4) } AB}
\]

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Final Summary of Answers:

1. \(\boxed{\frac{40}{3}}\)
2. \(\boxed{6}\)
3. \(\boxed{\frac{36}{5}}\)
4. \(\boxed{26}\)
5. \(\boxed{\text{1) } EG, \text{ 2) } CA, \text{ 3) } AG, \text{ 4) } AB}\)