Question

using the sss congruence theorem
triangle def and triangle dgf are shown in the diagram.
to prove that $\triangle def \cong \triangle dgf$ by sss, what additional information is needed?
options:
$\overline{de} \cong \overline{dg}$
$\overline{dg} \cong \overline{gf}$
$\angle dfe \cong \angle dfg$
$\angle def \cong \angle dgf$

Explanation:

Step1: Recall SSS Congruence

SSS (Side - Side - Side) congruence states that if three sides of one triangle are equal to three corresponding sides of another triangle, the triangles are congruent.

Step2: Analyze Given Information

From the diagram, we can see that \(EF = GF\) (marked with the same tick marks) and \(DF\) is a common side to both \(\triangle DEF\) and \(\triangle DGF\), so \(DF=DF\) (by the reflexive property).

Step3: Determine the Missing Side

For SSS congruence, we need the third pair of sides to be equal. The third sides of \(\triangle DEF\) and \(\triangle DGF\) are \(DE\) and \(DG\) respectively. So we need \(DE\cong DG\) to satisfy the SSS congruence criterion.

Let's analyze the other options:

• Option \(\overline{DG}\cong\overline{GF}\): We already know \(\overline{EF}\cong\overline{GF}\), and this does not help in proving \(\triangle DEF\cong\triangle DGF\) by SSS.
• Option \(\angle DFE\cong\angle DFG\): This is an angle - related condition, not a side - related condition for SSS.
• Option \(\triangle DEF\cong\triangle DGF\): This is what we are trying to prove, not the additional information needed.

Answer:

\(\overline{DE}\cong\overline{DG}\) (the option with the text \(\overline{DE}\cong\overline{DG}\))