Question

4 *

yes, sss
yes, sas
yes, aas
no

Explanation:

Step1: Analyze the given information

We have two triangles \(\triangle EMA\) and \(\triangle EVA\). We know that \(EM = EV\) (given as marked equal), \(\angle M=\angle V\) (given as marked equal), and \(EA = EA\) (common side).

Step2: Recall the congruence criteria

The SAS (Side - Angle - Side) congruence criterion states that if two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, then the two triangles are congruent. Here, the side \(EM = EV\), the angle \(\angle M=\angle V\), and the side \(EA = EA\) (but the angle is not included between the two pairs of equal sides for SSS. For AAS, we need two angles and a non - included side. Here we have two sides and a non - included angle. Wait, no, actually, if we consider the two triangles \(\triangle EMA\) and \(\triangle EVA\), we have \(EM = EV\), \(\angle M=\angle V\), and \(EA = EA\). The correct congruence criterion is SAS because we have two sides and the included angle (the angle between \(EM\) and \(EA\) in \(\triangle EMA\) and between \(EV\) and \(EA\) in \(\triangle EVA\) is not relevant. Wait, no, actually, if we consider the two triangles:
Let's re - express. In \(\triangle EMA\) and \(\triangle EVA\):

• \(EM = EV\) (given)
• \(\angle M=\angle V\) (given)
• \(EA = EA\) (common side)

The congruence criterion is SAS (the side - angle - side where the angle is between the two sides. Wait, no, hold on. Wait, actually, if we consider the two triangles \(\triangle EMA\) and \(\triangle EVA\), we have two sides \(EM = EV\), \(EA=EA\) and the non - included angle \(\angle M=\angle V\). But no, wait, no. Wait, actually, if we look at the triangles, the two sides \(EM = EV\), \(EA = EA\) and the angle \(\angle M=\angle V\). The correct congruence is SAS. Wait, no, hold on. Wait, the two triangles \(\triangle EMA\) and \(\triangle EVA\):
We have \(EM = EV\) (given), \(\angle M=\angle V\) (given), \(EA = EA\) (common side). The angle is between \(EM\) and \(MA\) in \(\triangle EMA\) and between \(EV\) and \(VA\) in \(\triangle EVA\). But we know \(EM = EV\), \(EA = EA\) (the two sides) and \(\angle M=\angle V\) (the included angle? No. Wait, no, wait, the two triangles:
Let’s use the SAS formula. For \(\triangle EMA\) and \(\triangle EVA\):

• \(EM = EV\) (side)
• \(\angle M=\angle V\) (angle)
• \(MA = VA\)? No, not given. Wait, no, wait, the figure shows that \(EM = EV\) (marked), \(EA\) is common. And \(\angle M=\angle V\). So by SAS (the two sides \(EM = EV\), \(EA = EA\) and the included angle (the angle between \(EM\) and \(EA\) in \(\triangle EMA\) and between \(EV\) and \(EA\) in \(\triangle EVA\) is not. Wait, no, actually, if we consider the two triangles \(\triangle EMA\) and \(\triangle EVA\):

We have two sides \(EM = EV\), \(EA = EA\) and the angle \(\angle M=\angle V\). The congruence is by SAS.

Answer:

yes, SAS