Question

geometry sem a - a triangle similarity theorems use the converse of the side - splitter theorem to determine if $overline{tu} parallel overline{rs}$. which statement is true?
line segment tu is not parallel to line segment rs because $\frac{32}{45}
eq \frac{40}{36}$.
line segment tu is parallel to line segment rs because $\frac{32}{36} = \frac{40}{45}$.
line segment tu is not parallel to line segment rs because $\frac{32}{36}
eq \frac{40}{45}$.
line segment tu is parallel to line segment rs because $\frac{32}{45} = \frac{40}{36}$.

Explanation:

Step 1: Recall the Side - Splitter Theorem (Converse)

The converse of the side - splitter theorem states that if a line divides two sides of a triangle proportionally, then the line is parallel to the third side. In triangle \( QRS \), the line \( TU \) divides \( QR \) and \( QS \) into segments \( QT = 32 \), \( TR=36 \), \( QU = 40 \), and \( US = 45 \).

Step 2: Check the Proportions

We need to check if \( \frac{QT}{TR}=\frac{QU}{US} \).
Calculate \( \frac{QT}{TR}=\frac{32}{36}=\frac{8}{9}\) (dividing numerator and denominator by 4).
Calculate \( \frac{QU}{US}=\frac{40}{45}=\frac{8}{9}\) (dividing numerator and denominator by 5).
Since \( \frac{32}{36}=\frac{40}{45}\) (both equal to \( \frac{8}{9}\)), by the converse of the side - splitter theorem, \( TU\parallel RS \).

Answer:

Line segment \( \overline{TU} \) is parallel to line segment \( \overline{RS} \) because \( \frac{32}{36} = \frac{40}{45} \).