Question

is \\( \triangle a b c \cong \triangle d f e \\) by aas theorem?
(1 point)
yes. the aas theorem is used.
no. the asa theorem is used
no. \\( \overline { d e } \\) does not correspond to \\( \overline { b c } \\).
yes \\( \overline { b c } \cong \overline { f e } \\)

Explanation:

Step1: Recall AAS and ASA theorems

AAS (Angle - Angle - Side) theorem states that if two angles and a non - included side of one triangle are congruent to two angles and the corresponding non - included side of another triangle, then the triangles are congruent. ASA (Angle - Side - Angle) theorem states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.
In \(\triangle ABC\) and \(\triangle DFE\), we have two angles (the angles at \(A\) and \(B\) in \(\triangle ABC\) and the angles at \(D\) and \(F\) in \(\triangle DFE\)) and a non - included side (\(BC\) and \(DE\)). But for AAS, the non - included sides should correspond. Here, \(BC\) and \(DE\) are not corresponding sides.

Answer:

No. \(\overline{DE}\) does not correspond to \(\overline{BC}\).