Question

1. in $\triangle aeb$, $overline{cd}$ is parallel to $overline{ab}$. complete the proportion to the right. (2 pts)
2. solve for x in the triangle below. (4 pts)
3. given: $overline{fk} parallel overline{gh} parallel overline{ij}$, solve for x. (4 pts)
4. $\triangle ghj$ is shown. is $overline{fi} parallel overline{hj}$? explain how you know. (4 pts)

Explanation:

Step1: Apply Triangle Proportionality Theorem

For $\triangle AEB$ with $\overline{CD} \parallel \overline{AB}$, the proportion is $\frac{EC}{CA} = \frac{ED}{DB}$ (or equivalent $\frac{EC}{EA} = \frac{ED}{EB}$, $\frac{CA}{EC} = \frac{DB}{ED}$, $\frac{CA}{EA} = \frac{DB}{EB}$)

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Step1: Set up proportionality equation

$\frac{x}{25} = \frac{8}{20}$

Step2: Solve for x

$x = 25 \times \frac{8}{20}$
$x = 10$

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Step1: Set up proportionality equation

$\frac{4}{12} = \frac{5}{x}$

Step2: Cross-multiply to solve x

$4x = 12 \times 5$
$x = \frac{60}{4}$
$x = 15$

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Step1: Calculate segment ratios

$\frac{GF}{FH} = \frac{5.0}{3.0} = \frac{5}{3}$
$\frac{GI}{IJ} = \frac{4.0}{2.4} = \frac{5}{3}$

Step2: Apply converse of theorem

Since $\frac{GF}{FH} = \frac{GI}{IJ}$, $\overline{FI} \parallel \overline{HJ}$

Answer:

1. $\boldsymbol{\frac{EC}{CA} = \frac{ED}{DB}}$ (any equivalent valid proportion is acceptable)
2. $\boldsymbol{10}$
3. $\boldsymbol{15}$
4. $\boldsymbol{\overline{FI} \parallel \overline{HJ}}$; The ratios of the divided segments on each side of $\triangle GHJ$ are equal ($\frac{GF}{FH} = \frac{GI}{IJ} = \frac{5}{3}$), so by the Converse of the Triangle Proportionality Theorem, the segments are parallel.