Question

in addition to the facts in the diagram, which other statements are necessary to prove that δabc is congruent to δefg by the asa criterion?
i. ( mangle b = mangle f )
ii. ( bc = fg )
iii. ( mangle a = mangle e )
iv. ( fg = 3 )
v. ( mangle b = mangle e )

Explanation:

Step1: Recall ASA Criterion

ASA (Angle - Side - Angle) congruence criterion 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 two triangles are congruent.

Step2: Analyze Given Diagram and Options

From the diagram, we can see that \(AB = EF = 2\) (one pair of sides) and \(\angle B\) and \(\angle E\) are marked as equal (one pair of angles). For ASA, we need another pair of angles such that the side is included between the two angles. So we need \(\angle A=\angle E\) (wait, no, let's re - examine). Wait, the side \(AB = EF\). For ASA, we need two angles and the included side. Let's look at the triangles \(\triangle ABC\) and \(\triangle EFG\). We know \(AB = EF = 2\). We have \(\angle B\) and \(\angle E\) (the marked angles) as equal. For ASA, we need the angle at \(A\) and angle at \(E\) to be equal (so that the side \(AB\) is between \(\angle A\) and \(\angle B\) in \(\triangle ABC\) and side \(EF\) is between \(\angle E\) and \(\angle F\) in \(\triangle EFG\))? Wait, no, let's check the options:

• Option i: \(m\angle B=m\angle F\): But we already have \(\angle B=\angle E\) (marked), so this is not for ASA.
• Option ii: \(BC = FG\): This is a side, not an angle, so not for ASA.
• Option iii: \(m\angle A=m\angle E\): If \(\angle A=\angle E\), \(AB = EF\), and \(\angle B=\angle E\)? Wait, no, \(\angle B\) and \(\angle E\) are the marked angles. Wait, actually, in \(\triangle ABC\), the included side between \(\angle A\) and \(\angle B\) is \(AB\). In \(\triangle EFG\), the included side between \(\angle E\) and \(\angle F\) is \(EF\). We know \(AB = EF\). If we have \(\angle A=\angle E\) and \(\angle B=\angle F\)? No, wait the marked angle is \(\angle B\) in \(\triangle ABC\) and \(\angle E\) in \(\triangle EFG\). Wait, maybe I made a mistake. Let's re - express:

Wait, the correct approach: For ASA, we need two angles and the included side. Let's assume that in \(\triangle ABC\) and \(\triangle EFG\), we have:

• We know \(AB = EF\) (length 2).
• We need \(\angle A=\angle E\) and \(\angle B=\angle F\)? No, the marked angle is \(\angle B\) (in \(\triangle ABC\)) and \(\angle E\) (in \(\triangle EFG\)). Wait, maybe the marked angle is \(\angle B\) and \(\angle E\) are equal. Then, for ASA, we need the angle at \(A\) (in \(\triangle ABC\)) and angle at \(E\) (in \(\triangle EFG\)) to be equal, and the side \(AB\) (between \(\angle A\) and \(\angle B\)) and \(EF\) (between \(\angle E\) and \(\angle F\)) to be equal. So if \(m\angle A=m\angle E\), then we have \(\angle A=\angle E\), \(AB = EF\), and \(\angle B=\angle E\)? No, \(\angle B\) and \(\angle E\) are the marked angles. Wait, maybe the marked angle is \(\angle B\) in \(\triangle ABC\) and \(\angle E\) in \(\triangle EFG\) are equal. Then, if \(m\angle A=m\angle E\) (wait, no, \(m\angle A=m\angle E\)) and \(AB = EF\), and \(\angle B=\angle E\) (marked), then by ASA, \(\triangle ABC\cong\triangle EFG\).
• Option iv: \(FG = 3\): This is a side length, not an angle, so not for ASA.
• Option v: \(m\angle B=m\angle E\): We already can see from the diagram that \(\angle B\) and \(\angle E\) are marked as equal, so this is not an additional statement.

So the correct option is iii: \(m\angle A=m\angle E\)

Answer:

iii. \(m\angle A = m\angle E\)