Question

find uv.
uv = \boxed{}

Explanation:

Step1: Identify the Midline Theorem

The segment \( XU \) is parallel to \( WV \) (indicated by the arrows), so by the Midline Theorem (or Basic Proportionality Theorem), the ratio of \( WX \) to \( XT \) should equal the ratio of \( UV \) to \( TU \). Wait, actually, since \( XU \) is a midline? Wait, no, let's check the lengths. \( WX = 14 \), \( XT = 28 \), so \( WT = WX + XT = 14 + 28 = 42 \)? Wait, no, \( WX = 14 \), \( XT = 28 \), so the ratio of \( WX \) to \( WT \) is \( \frac{14}{14 + 28} = \frac{14}{42} = \frac{1}{3} \)? Wait, no, maybe \( X \) divides \( WT \) into \( WX = 14 \) and \( XT = 28 \), so \( WX:XT = 14:28 = 1:2 \). Then, by the Basic Proportionality Theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. Here, \( XU \parallel WV \), so \( \frac{WX}{WT} = \frac{UV}{TV} \)? Wait, no, \( TV = TU + UV = 44 + UV \), and \( WT = WX + XT = 14 + 28 = 42 \). Wait, maybe I got the segments wrong. Let's re-examine the triangle. The triangle is \( \triangle WTV \), with a line \( XU \) parallel to \( WV \), intersecting \( WT \) at \( X \) and \( TV \) at \( U \). So by Thales' theorem, \( \frac{WX}{WT} = \frac{VU}{TV} \)? Wait, no, \( \frac{WX}{XT} = \frac{UV}{TU} \)? Wait, \( WX = 14 \), \( XT = 28 \), so \( \frac{WX}{XT} = \frac{14}{28} = \frac{1}{2} \). Then, since \( XU \parallel WV \), \( \frac{UV}{TU} = \frac{WX}{XT} = \frac{1}{2} \). Wait, \( TU = 44 \), so \( \frac{UV}{44} = \frac{1}{2} \)? Wait, that would make \( UV = 22 \). Wait, let's check again. Wait, maybe \( X \) is the midpoint? No, \( WX = 14 \), \( XT = 28 \), so \( XT = 2 \times WX \). So the ratio of \( WX \) to \( WT \) is \( 14:(14 + 28) = 1:3 \), but that doesn't seem right. Wait, maybe the line \( XU \) is parallel to \( WV \), so \( \triangle WXV \) and \( \triangle TXU \) are similar? Wait, no, \( XU \parallel WV \), so \( \triangle TXU \sim \triangle TWV \) by AA similarity (since \( \angle TXU = \angle TWV \) and \( \angle T \) is common). Therefore, the ratio of corresponding sides is equal. So \( \frac{TX}{TW} = \frac{TU}{TV} \). \( TX = 28 \), \( TW = TX + XW = 28 + 14 = 42 \), \( TU = 44 \), \( TV = TU + UV = 44 + UV \). So \( \frac{28}{42} = \frac{44}{44 + UV} \). Simplify \( \frac{28}{42} = \frac{2}{3} \). So \( \frac{2}{3} = \frac{44}{44 + UV} \). Cross-multiplying: \( 2(44 + UV) = 3 \times 44 \). \( 88 + 2UV = 132 \). Subtract 88: \( 2UV = 44 \). Divide by 2: \( UV = 22 \). Wait, that makes sense. Let's verify. If \( UV = 22 \), then \( TV = 44 + 22 = 66 \). Then \( \frac{TX}{TW} = \frac{28}{42} = \frac{2}{3} \), and \( \frac{TU}{TV} = \frac{44}{66} = \frac{2}{3} \), so the ratios are equal, confirming similarity. So that works.

Step2: Apply Thales' Theorem (Similar Triangles)

We have \( \triangle TXU \sim \triangle TWV \) (by AA similarity, as \( XU \parallel WV \), so corresponding angles are equal). Therefore, the ratio of corresponding sides is equal:

$$ \frac{TX}{TW} = \frac{TU}{TV} $$

Substitute the known values: \( TX = 28 \), \( TW = TX + XW = 28 + 14 = 42 \), \( TU = 44 \), and \( TV = TU + UV = 44 + UV \).

$$ \frac{28}{42} = \frac{44}{44 + UV} $$

Simplify \( \frac{28}{42} = \frac{2}{3} \):

$$ \frac{2}{3} = \frac{44}{44 + UV} $$

Cross-multiply to solve for \( UV \):

$$ 2(44 + UV) = 3 \times 44 $$

$$ 88 + 2UV = 132 $$

Subtract 88 from both sides:

$$ 2UV = 132 - 88 = 44 $$

Divide by 2:

$$ UV = \frac{44}{2} = 22 $$

Answer:

\( 22 \)