5)
Consider triangle FEC and triangle FED
Given that,
[tex]\begin{gathered} \bar{FC}=\bar{FD} \\ \bar{CE}=\bar{DE} \end{gathered}[/tex]Also,
[tex]\bar{FE}=\bar{FE}\text{ (common)}[/tex]Therefore, by, SSS congruence rule,
[tex]\Delta FEC\text{ }\cong\Delta FED[/tex]6)
Consider triangle DAJ and triangle EAM
Given that,
[tex]\begin{gathered} \bar{DA}=\bar{EA} \\ \bar{DJ}=\bar{EM} \end{gathered}[/tex]Since linesegments, DJ parallel to ME,
[tex]\angle ADJ\text{ =}\angle AEM\text{ ( since, interior opposite angles are equal)}[/tex]Therefore, by SAS congruence rule,
[tex]\Delta DAJ\text{ }\cong\Delta\text{EAM}[/tex]