Question

identifying information necessary for applying the sss similarity theorem 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 - \triangle p q r \\) using the sss similarity theorem? \\( d e = p q \\) \\( \angle d = \angle p \\) \\( \frac { d e } { e f } = \frac { 3 } { 2 } \\) \\( \frac { d e } { p q } = \frac { 3 } { 2 } \\)

Explanation:

Step1: Recall SSS similarity theorem

The SSS (Side - Side - Side) similarity theorem states that if the ratios of the corresponding sides of two triangles are equal, then the two triangles are similar. For \(\triangle DEF\) and \(\triangle PQR\), we already have \(\frac{DF}{PR}=\frac{FE}{RQ}=\frac{3}{2}\). We need the ratio of the third pair of corresponding sides (\(DE\) and \(PQ\)) to also be \(\frac{3}{2}\).

Step2: Analyze each option

• Option \(DE = PQ\): If \(DE = PQ\), then \(\frac{DE}{PQ}=1

eq\frac{3}{2}\), so this is incorrect.

• Option \(\angle D=\angle P\): This is information for the SAS (Side - Angle - Side) similarity criterion, not SSS.
• Option \(\frac{DE}{EF}=\frac{3}{2}\): This ratio is between sides of \(\triangle DEF\) and not between corresponding sides of \(\triangle DEF\) and \(\triangle PQR\).
• Option \(\frac{DE}{PQ}=\frac{3}{2}\): This gives the ratio of the third pair of corresponding sides. When \(\frac{DF}{PR}=\frac{FE}{RQ}=\frac{DE}{PQ}=\frac{3}{2}\), by the SSS similarity theorem, \(\triangle DEF\sim\triangle PQR\).

Answer:

\(\frac{DE}{PQ}=\frac{3}{2}\) (the fourth option)