Question

select the correct answer from each drop - down menu.the conditionproves that \\( \triangle a b \\)\\( \triangle e g \\) are congruent by the sas criterion.if \\( a b \
eq e f \\), theis violated. in this situation, angle \\( c \\)angle \\( g \\).

Explanation:

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

For two triangles to be congruent by SAS, two sides and the included angle of one triangle must be equal to the corresponding two sides and included angle of the other triangle. In \(\triangle ABC\) and \(\triangle EFG\), we know \(AC = EG\) (marked as equal) and \(BC=FG\) (marked as equal). The included angles for these sides in \(\triangle ABC\) and \(\triangle EFG\) are \(\angle C\) and \(\angle G\) respectively. But for SAS, if we consider the sides \(AB\) and \(EF\) as part of the SAS condition (assuming the sides adjacent to the angles), we need \(AB = EF\) as the third - side (along with the two pairs of equal sides and the included angle concept, here the sides \(AC - EG\), \(BC - FG\) and if \(AB = EF\) then the triangles can be compared for SAS).

Step2: Analyze the non - congruence case

If \(AB
eq EF\), the SSS (Side - Side - Side) or SAS (if we assume the side - angle - side structure) condition (where all corresponding sides are equal for SSS and for SAS two sides and included angle) is violated. Using the law of cosines \(c^{2}=a^{2}+b^{2}-2ab\cos C\) (where in \(\triangle ABC\), \(AB^{2}=AC^{2}+BC^{2}-2(AC)(BC)\cos C\) and in \(\triangle EFG\), \(EF^{2}=EG^{2}+FG^{2}-2(EG)(FG)\cos G\)). Since \(AC = EG\) and \(BC = FG\), if \(AB
eq EF\), then \(\cos C
eq\cos G\). And since \(0 < C<\pi\) and \(0 < G<\pi\), \(C
eq G\)

Answer:

The first blank: \(AB = EF\); The second blank: is not congruent to