Question

st || uw. find uv.
39
t u v
w
66
22
s
uv =

Explanation:

Step1: Identify Similar Triangles

Since \( \overline{ST} \parallel \overline{UW} \), triangles \( \triangle STV \) and \( \triangle UWV \) are similar by the Basic Proportionality Theorem (Thales' theorem). So, the ratios of corresponding sides are equal. Let \( UV = x \), then \( TV = TU + UV = 39 + x \)? Wait, no, looking at the diagram: \( TV \) is 39? Wait, the top segment is \( TU + UV \)? Wait, no, the diagram shows \( T \) to \( V \) with \( U \) in between? Wait, no, the length from \( T \) to the top is 39, and the side length is 66 (from \( V \) to \( S \)?) Wait, no, the vertical side: \( VW = 22 \), \( VS = 66 \)? Wait, no, the right side has length 66, and the segment from \( W \) to \( S \) is 22, so \( VW = 66 - 22 = 44 \)? Wait, no, let's re-express.

Wait, the triangles are similar, so \( \frac{UV}{ST} = \frac{VW}{VS} \)? Wait, no, \( ST \parallel UW \), so \( \triangle TSV \sim \triangle USW \)? Wait, maybe the correct proportion is \( \frac{UV}{TV} = \frac{VW}{VS} \). Wait, let's define: Let \( UV = x \), \( TV = 39 \) (wait, the top segment is \( TV = 39 \)? No, the top segment is \( T \) to the right end, length 39, with \( U \) and \( V \) on it. So \( TU + UV = 39 \)? Wait, no, the diagram: \( T \), \( U \), \( V \) are colinear on the top, with \( T \) to \( V \) being 39? No, the top bar is labeled 39, so \( TV = 39 \). Then the vertical side: from \( V \) down to \( S \) is 66, and from \( W \) down to \( S \) is 22, so \( VW = 66 - 22 = 44 \)? Wait, no, \( VS = 66 \), \( WS = 22 \), so \( VW = VS - WS = 66 - 22 = 44 \). Now, since \( ST \parallel UW \), triangles \( \triangle UVW \) and \( \triangle STV \) are similar. So the ratio of sides: \( \frac{UV}{ST} = \frac{VW}{VS} \)? Wait, no, \( ST \) is parallel to \( UW \), so \( \angle T = \angle U \) (corresponding angles), \( \angle V \) is common. So \( \triangle UVW \sim \triangle STV \) by AA similarity. Therefore, \( \frac{UV}{TV} = \frac{VW}{VS} \). Wait, \( TV = 39 \), \( VW = 22 \), \( VS = 66 \)? No, \( VS = 66 \), \( WS = 22 \), so \( VW = VS - WS = 66 - 22 = 44 \)? Wait, no, maybe \( WS = 22 \), \( VS = 66 \), so \( \frac{VW}{VS} = \frac{22}{66} = \frac{1}{3} \)? No, that can't be. Wait, maybe the correct proportion is \( \frac{UV}{TV - UV} = \frac{VW}{WS} \)? No, let's start over.

Wait, the key is that \( ST \parallel UW \), so by the Basic Proportionality Theorem (Thales' theorem), \( \frac{UV}{TU} = \frac{VW}{WS} \). Wait, but we need to know \( TU \) or \( UV \). Wait, maybe \( TV = 39 \), \( VS = 66 \), \( WS = 22 \). So \( \frac{UV}{39 - UV} = \frac{22}{66 - 22} \)? No, this is getting confusing. Wait, the standard problem like this: when a line is parallel to a side of a triangle, it divides the other two sides proportionally. So if \( UW \parallel ST \), then in triangle \( TSV \), line \( UW \) is parallel to \( ST \), so \( \frac{UV}{TV} = \frac{UW}{ST} \)? No, maybe the correct proportion is \( \frac{UV}{TV} = \frac{VW}{VS} \). Wait, let's use variables. Let \( UV = x \), \( TV = 39 \), so \( TU = 39 - x \). Then, since \( UW \parallel ST \), \( \frac{TU}{UV} = \frac{WS}{VW} \). \( WS = 22 \), \( VW = 66 - 22 = 44 \). So \( \frac{39 - x}{x} = \frac{22}{44} = \frac{1}{2} \). Then \( 2(39 - x) = x \), \( 78 - 2x = x \), \( 78 = 3x \), \( x = 26 \). Wait, that makes sense. Let's check: \( TU = 39 - 26 = 13 \), \( \frac{TU}{UV} = \frac{13}{26} = \frac{1}{2} \), and \( \frac{WS}{VW} = \frac{22}{44} = \frac{1}{2} \). So the proportion holds. Therefore, \( UV = 26 \).

Step1: Set Up Proportion

Since \( ST \parallel UW \), by the Basic Proportionality Theorem (Thales' theorem), \( \frac{TU}{UV} = \frac{WS}{VW} \). Let \( UV = x \), then \( TU = 39 - x \). \( WS = 22 \), \( VW = 66 - 22 = 44 \). So:
\[ \frac{39 - x}{x} = \frac{22}{44} \]

Step2: Simplify and Solve

Simplify \( \frac{22}{44} = \frac{1}{2} \), so:
\[ \frac{39 - x}{x} = \frac{1}{2} \]
Cross-multiplying:
\[ 2(39 - x) = x \]
\[ 78 - 2x = x \]
\[ 78 = 3x \]
\[ x = 26 \]

Answer:

\( \boxed{26} \)