Question

completing a proof
given: \\( \triangle stu \\) with \\( \overline{st} \parallel \overline{xy} \\)
prove: \\( \frac{sx}{xu}=\frac{ty}{yu} \\)

complete the steps of the proof.

Explanation:

Step1: Identify similar triangles

Since \(\angle1=\angle2\) and \(\angle3 = \angle4\), by the AA (Angle - Angle) similarity theorem, \(\triangle UXY\sim\triangle UST\).

Step2: Use the property of similar triangles

For similar triangles \(\triangle UXY\) and \(\triangle UST\), the ratios of corresponding sides are equal. That is \(\frac{SU}{XU}=\frac{TU}{YU}\).

Step3: Apply segment addition

We know that \(SU = SX+XU\) and \(TU=TY + YU\) (segment addition postulate). Substituting these into \(\frac{SU}{XU}=\frac{TU}{YU}\), we get \(\frac{SX + XU}{XU}=\frac{TY+YU}{YU}\).

Step4: Simplify the equation

Using the rule \(\frac{a + b}{b}=\frac{a}{b}+1\), \(\frac{SX+XU}{XU}=\frac{SX}{XU}+ 1\) and \(\frac{TY + YU}{YU}=\frac{TY}{YU}+1\). Then \(\frac{SX}{XU}+1=\frac{TY}{YU}+1\). Subtracting \(1\) from both sides gives \(\frac{SX}{XU}=\frac{TY}{YU}\).

Answer:

\(\frac{SX}{XU}=\frac{TY}{YU}\)