Question

given that \\( \frac { d f } { p r } = \frac { f e } { r q } = \frac { 3 } { 2 } \\), what additional information is needed to prove \\( \triangle d e f \sim \triangle p q r \\) using the sss similarity theorem?
\\( \bigcirc d e \cong p q \\)
\\( \bigcirc \angle d \cong \angle p \\)
\\( \bigcirc \frac { d e } { e f } = \frac { 3 } { 2 } \\)
\\( \bigcirc \frac { d e } { p q } = \frac { 3 } { 2 } \\)

Explanation:

Step1: Recall SSS similarity theorem

SSS similarity theorem states that if \(\frac{AB}{A'B'}=\frac{BC}{B'C'}=\frac{AC}{A'C'}\), then \(\triangle ABC\sim\triangle A'B'C'\)

Step2: Analyze given ratios

We are given \(\frac{DF}{PR}=\frac{FE}{RQ}=\frac{3}{2}\). For \(\triangle DEF\sim\triangle PQR\) by SSS similarity, we need the ratio of the third pair of corresponding sides \(DE\) and \(PQ\) to also be \(\frac{3}{2}\), i.e., \(\frac{DE}{PQ}=\frac{3}{2}\)

• \(DE\cong PQ\) gives a ratio of \(1\) (not \(\frac{3}{2}\)), so it's incorrect.
• \(\angle D\cong\angle P\) is used for SAS similarity (not SSS), so it's incorrect.
• \(\frac{DE}{EF}=\frac{3}{2}\) is a ratio of sides within \(\triangle DEF\) (not corresponding sides of \(\triangle DEF\) and \(\triangle PQR\)), so it's incorrect.

Answer:

\(\frac{DE}{PQ}=\frac{3}{2}\)