Answer with explanation:
[tex]\bar{AB}\cong\bar{CD}[/tex]
---------------------[Given]
[tex]\bar{AB}=\bar{CD}[/tex]
-----------------[By definition of congruent segments]---(2)
Also, AB= AE + EB
CD= CF +FD
Substituting the value of , AB and CD in equation (2),which reduces to
A E + E B=CF +FD
Also, [tex]\bar{CF}=\bar{EB}[/tex]
CF=E B--------[By definition of congruent segments]
A E + E B=FD +E B→→[Substitution Property]
Subtracting EB , from both sides
AE +EB - EB = FD + EB - EB
Which gives, A E = F D→→→→[Subtraction Property of Equality]
[tex]\bar{AE}\cong\bar{FD}[/tex]
→→→The Missing reason in the proof is :
Option(D.): Subtraction Property of Equality
