What additional information can be used to prove δfeg ≅ δfqg by sas?

∠EFG=∠QFG

Step-by-step explanation:

The SAS postulate of congruence says that if two sides and the included angle of a triangle are congruent to two sides and the included angle of another triangle, then these two triangles must be congruent.

In the given picture we have two triangles ΔFEG and ΔFQG  in which EF=FQ

and GF=GF [ by reflexive property].

To apply SAS postulate we need included angle of sides EF, GF and FQ, GF.

Therefore, the additional information can be used to prove ΔFEG ≅ ΔFQG is given by :-

∠EFG=∠QFG

Explanation:

In this exercise we have the following function:

So the domain of this function is the set of all real numbers such that

Why?

Remember that the domain of a function are those x-values for which the function is defined. In this case, for the function is undefined, therefore x can't be 5.