Question

given: △stu with \\(\overline{st} \parallel \overline{xy}\\)
prove: \\(\frac{sx}{xu} = \frac{ty}{yu}\\)
complete the steps of the proof.
diagram: triangle u with vertices y (on ut), t; x (on us), s. angles: ∠1 at y, ∠2 at t; ∠3 at x, ∠4 at s, with ∠1 and ∠2, ∠3 and ∠4 marked as corresponding angles (red arcs).
statements | reasons
--- | ---

1. △stu with \\(\overline{st} \parallel \overline{xy}\\) | 1. given
2. ∠1 and ∠2 are corresponding angles | 2. def. of corresponding angles
3. ∠3 and ∠4 are corresponding angles | 3. def. of corresponding angles
4. ∠1 ≅ ∠2; ∠3 ≅ ∠4 | 4. corresponding angles theorem
5. ♣ | 5. aa similarity theorem
6. \\(\frac{su}{xu} = \frac{tu}{yu}\\) | 6. def. of similar triangles
7. \\(su = sx + xu\\); \\(tu = ty + yu\\) | 7. segment addition postulate
8. \\(\frac{sx + xu}{xu} = \frac{ty + yu}{yu}\\) | 8. substitution property

dropdown options: substitution property, subtraction property, transitive property

Explanation:

Step 1: Analyze Step 5 (AA Similarity Theorem)

The AA (Angle - Angle) similarity theorem states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. From steps 2 - 4, we have $\angle1\cong\angle2$ and $\angle3\cong\angle4$. So the two triangles that must be similar are $\triangle UXY$ and $\triangle US T$. So the statement for step 5 should be $\triangle UXY\sim\triangle US T$ (by AA similarity theorem, since two pairs of corresponding angles are congruent).

Step 2: Analyze the Last Step (After Step 8)

We start with $\frac{SU}{XU}=\frac{TU}{YU}$ (from step 6) and $SU = SX+XU$, $TU=TY + YU$ (from step 7). Then we substitute $SU$ and $TU$ in the proportion: $\frac{SX + XU}{XU}=\frac{TY+YU}{YU}$. We can rewrite this fraction as $\frac{SX}{XU}+\frac{XU}{XU}=\frac{TY}{YU}+\frac{YU}{YU}$. Since $\frac{XU}{XU} = 1$ and $\frac{YU}{YU}=1$, we get $\frac{SX}{XU}+ 1=\frac{TY}{YU}+1$. Then, using the subtraction property of equality (subtract 1 from both sides), we have $\frac{SX}{XU}=\frac{TY}{YU}$. So the property used in the last step (to go from $\frac{SX + XU}{XU}=\frac{TY + YU}{YU}$ to $\frac{SX}{XU}=\frac{TY}{YU}$) is the subtraction property of equality (subtract 1 from both sides, and since $\frac{XU}{XU}=1$ and $\frac{YU}{YU} = 1$).

For step 5:

Step 1: Recall AA Similarity

AA similarity needs two congruent angle pairs. We have $\angle1\cong\angle2$ and $\angle3\cong\angle4$, so $\triangle UXY\sim\triangle UST$.
<Expression>$\triangle UXY\sim\triangle UST$</Expression>

For the last step (property):

Step 1: Simplify the Fractions

$\frac{SX + XU}{XU}=\frac{SX}{XU}+\frac{XU}{XU}=\frac{SX}{XU}+1$ and $\frac{TY + YU}{YU}=\frac{TY}{YU}+\frac{YU}{YU}=\frac{TY}{YU}+1$.

Step 2: Apply Subtraction Property

Subtract 1 from both sides: $\frac{SX}{XU}+1 - 1=\frac{TY}{YU}+1 - 1$, so $\frac{SX}{XU}=\frac{TY}{YU}$. The property is subtraction property.
<Expression>subtraction property</Expression>

Final Answers (for the blanks)

• Step 5 statement: $\boldsymbol{\triangle UXY\sim\triangle UST}$
• Last step property: $\boldsymbol{subtraction\ property}$

Answer:

Step 1: Simplify the Fractions

$\frac{SX + XU}{XU}=\frac{SX}{XU}+\frac{XU}{XU}=\frac{SX}{XU}+1$ and $\frac{TY + YU}{YU}=\frac{TY}{YU}+\frac{YU}{YU}=\frac{TY}{YU}+1$.

Step 2: Apply Subtraction Property

Subtract 1 from both sides: $\frac{SX}{XU}+1 - 1=\frac{TY}{YU}+1 - 1$, so $\frac{SX}{XU}=\frac{TY}{YU}$. The property is subtraction property.
<Expression>subtraction property</Expression>

Final Answers (for the blanks)

• Step 5 statement: $\boldsymbol{\triangle UXY\sim\triangle UST}$
• Last step property: $\boldsymbol{subtraction\ property}$