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prove: $\\triangle gec \\cong \\triangle hfa$.\ step\\tstatement\\treas…

Question

prove: $\triangle gec \cong \triangle hfa$.\
step\tstatement\treason\
\\t$abcd$ is a parallelogram\\t\
1\\t$\angle gec \cong \angle hfa$\tgiven\
\\t$\overline{ae} \cong \overline{fc}$\\t\
2\\t$\overline{ef} \cong \overline{ef}$\treflexive property\
3\\t$\overline{af} \cong \overline{ce}$\tsubstitution\
4\\t$\overline{bc} \parallel \overline{ad}$\tselect a reason...

Explanation:

Step1: Recall Parallelogram Property

In a parallelogram, opposite sides are parallel. Since \(ABCD\) is a parallelogram, \(BC \parallel AD\) by the definition of a parallelogram (opposite sides of a parallelogram are parallel).

Step2: Confirm the Reason

The reason for the statement \(\overline{BC} \parallel \overline{AD}\) is that "Opposite sides of a parallelogram are parallel" (or "Definition of a parallelogram").

Answer:

The reason for \(\overline{BC} \parallel \overline{AD}\) is "Opposite sides of a parallelogram are parallel" (or "Definition of a parallelogram").