Question

find tu.
w, v, u are on a line segment with vu = 23. s is a vertex connected to w, v, u. the length from w to the top horizontal line is 69. the length from s to u (passing through t) is 44 with t between s and u. there is a red arrow on ws and a red arrow on vt. we need to find tu and there is a box for tu =.

Explanation:

Step1: Identify the midline theorem

The segment \( TV \) is parallel to \( WS \) (implied by the arrows, indicating midlines or parallel lines), so by the Basic Proportionality Theorem (Thales' theorem), the ratio of \( UV \) to \( UW \) should be equal to the ratio of \( UT \) to \( US \). First, find the length of \( WV \): \( WV = WU - UV = 69 - 23 = 46 \). Wait, actually, since \( TV \) is a midline or parallel, the triangles \( TUV \) and \( S UW \) are similar. So the ratio of \( UV \) to \( UW \) is \( \frac{23}{69}=\frac{1}{3} \). Then, the ratio of \( TU \) to \( SU \) should also be \( \frac{1}{3} \). Let \( TU = x \), then \( SU = ST + TU = 44 + x \). So \( \frac{x}{44 + x}=\frac{23}{69}=\frac{1}{3} \).

Step2: Solve the proportion

Cross - multiply: \( 3x=44 + x \). Subtract \( x \) from both sides: \( 3x - x=44 + x - x \), which gives \( 2x = 44 \). Then divide both sides by 2: \( x=\frac{44}{2}=22 \). Wait, another way: Since \( \frac{UV}{WU}=\frac{23}{69}=\frac{1}{3} \), and because of similar triangles, \( \frac{TU}{SU}=\frac{1}{3} \). Let \( TU = x \), \( SU=44 + x \), so \( 3x=44 + x \), \( 2x = 44 \), \( x = 22 \). Or, notice that \( 69\div23 = 3 \), so the scale factor is \( \frac{1}{3} \). Then \( TU=\frac{44}{2}=22 \) (since the ratio of sides in similar triangles is \( 1:3 \), so the smaller triangle's side \( TU \) and the larger triangle's side \( SU \) have a ratio of \( 1:3 \), and \( SU - TU=44 \), let \( TU = x \), \( SU = 3x \), then \( 3x - x = 44 \), \( 2x = 44 \), \( x = 22 \)).

Answer:

\( 22 \)