Respuesta :

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SOLUTION

[tex]\begin{gathered} \Delta FED\cong\Delta CBA \\ \angle F\cong\angle C \end{gathered}[/tex][tex]\begin{gathered} \Delta EFD\cong BCA \\ \angle E\cong\angle B \\ \bar{DE}\cong\bar{AB} \end{gathered}[/tex][tex]\begin{gathered} \bar{EF}=\bar{BC} \\ \angle D=\angle A \\ \bar{DF}=\bar{AC} \end{gathered}[/tex]

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