Answer:
c
Step-by-step explanation:
Answer:
The one with arrows are the answers
->Line segment E B is bisected by Line segment D F .
->A is the midpoint of Line segment F C .
Line segment F C bisects Line segment D B.
->Line segment E B is a segment bisector.
->FA = One-halfFC.
Line segment D A is congruent to Line segment A B .
Step-by-step explanation:
I did it on edge and got it right
Answer:
Prove set equality by showing that for any element 
, 
if and only if 
.
Example:
.
.
.
.
.
Step-by-step explanation:
Proof for ![[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]](https://tex.z-dn.net/?f=%5Bx%20%5Cin%20%28A%20%5Cbackslash%20%28B%20%5Ccap%20C%29%29%5D%20%5Cimplies%20%5Bx%20%5Cin%20%28%28A%20%5Cbackslash%20B%29%20%5Ccup%20%28A%20%5Cbackslash%20C%29%29%5D)
for any element :
Assume that . Thus, 
and 
.
Since , either 
or 
(or both.)
- If , then combined with ,
. - Similarly, if , then combined with ,
.
Thus, either or (or both.)
Therefore, as required.
Proof for ![[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]](https://tex.z-dn.net/?f=%5Bx%20%5Cin%20%28%28A%20%5Cbackslash%20B%29%20%5Ccup%20%28A%20%5Cbackslash%20C%29%29%5D%20%5Cimplies%20%5Bx%20%5Cin%20%28A%20%5Cbackslash%20%28B%20%5Ccap%20C%29%29%5D)
:
Assume that . Thus, either or (or both.)
- If , then and . Notice that
since the contrapositive of that statement, 
, is true. Therefore, and thus 
. - Otherwise, if
, then and . Similarly, 
implies . Therefore, .
Either way, .
Therefore, implies , as required.
