Question

which of the following is not sufficient to prove two triangles are congruent by sss?
a. ab = de, bc = df, ac = ef where sides don’t correspond
b. ab = de, ac = df, bc = ef
c. ab = de, ac = df, bc = ef where sides correspond correctly
d. ab = bc, ac = ef, bc = de

Explanation:

Step1: Recall SSS congruence rule

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

Step2: Analyze option a

In option a, \(AB = DE\), \(BC=DF\), \(AC = EF\). Here, the sides do not correspond correctly. For SSS, the sides should be in the order of corresponding vertices. If we consider \(\triangle ABC\) and \(\triangle DEF\), the correct correspondence for SSS would be \(AB = DE\), \(BC=EF\), \(AC = DF\).

Step3: Analyze option b

In option b, \(AB = DE\), \(AC = DF\), \(BC=EF\). Here, \(AB\) corresponds to \(DE\), \(AC\) corresponds to \(DF\), and \(BC\) corresponds to \(EF\). So, by SSS, \(\triangle ABC\cong\triangle DEF\).

Step4: Analyze option c

In option c, \(AB = DE\), \(AC = DF\), \(BC = EF\). Here, \(AB\) corresponds to \(DE\), \(AC\) corresponds to \(DF\), and \(BC\) corresponds to \(EF\). So, by SSS, \(\triangle ABC\cong\triangle DEF\).

Step5: Analyze option d

In option d, \(AB = BC\), \(AC = EF\), \(BC = DE\). We can rewrite it as \(AB = BC\), \(BC = DE\), \(AC = EF\). If \(AB = BC\) and \(BC = DE\), then \(AB = DE\). Now we have \(AB = DE\), \(AC = EF\), \(BC = DE\) (but \(BC = DE\) and \(AB = DE\) implies \(AB = BC\)). If we consider two triangles (say \(\triangle ABC\) and \(\triangle DEF\)), we can assume a correspondence and check SSS.

Answer:

A. \(AB = DE, BC = DF, AC = EF\) where sides don't correspond