Question

select the correct answer from each drop - down menu.
the condition proves that (\triangle abc) and (\triangle efg) are congruent by the sas criterion.
if (ab
eq ef), the is violated. in this situation, angle (c) angle (g).

Explanation:

Step1: Recall the SAS (Side - Angle - Side) congruence criterion

For two triangles \(\triangle ABC\) and \(\triangle EFG\) to be congruent by SAS, we need two sides and the included angle of one triangle to be equal to the corresponding two sides and included angle of the other triangle.
We know that \(CB = FG\) (marked with two short lines) and \(AC=EG\) (marked with one short line). So, for SAS, we need the included angles. The included angle for sides \(AC\) and \(CB\) in \(\triangle ABC\) is \(\angle B\), and the included angle for sides \(EG\) and \(FG\) in \(\triangle EFG\) is \(\angle F\). So, \(\angle B\cong\angle F\) is the condition for SAS.

Step2: Analyze the non - congruence case

If \(AB
eq EF\), the SAS criterion (since we have two sides \(AC = EG\), \(CB=FG\) and the included angles \(\angle B=\angle F\)) is violated.

Step3: Use the property of non - congruent triangles (but with two sides equal)

We use the Law of Cosines. In \(\triangle ABC\), \(AB^{2}=AC^{2}+CB^{2}-2(AC)(CB)\cos C\). In \(\triangle EFG\), \(EF^{2}=EG^{2}+FG^{2}-2(EG)(FG)\cos G\). Since \(AC = EG\) and \(CB = FG\), if \(AB
eq EF\), then \(\cos C
eq\cos G\), so \(\angle C
eq\angle G\)

Answer:

The first blank: angle \(B\) is congruent to angle \(F\); the second blank: SAS criterion; the third blank: is not congruent to