Question

use the given information to prove that △def ≅ △fgd
given: \overline{de} ≅ \overline{fg}
\overline{ef} ≅ \overline{gd}
prove: △def ≅ △fgd

Explanation:

Step1: Fill in the first blank

Given that \(\overline{DE}\cong\overline{FG}\), so the first blank is \(FG\).

Step2: State the reason for \(\overline{EF}\cong\overline{GD}\)

The reason for \(\overline{EF}\cong\overline{GD}\) is "Given".

Step3: Fill in the third blank

By the reflexive property of congruence, \(\overline{DF}\cong\overline{DF}\), so the third blank is \(DF\).

Step4: State the reason for \(\triangle DEF\cong\triangle FGD\)

Since we have \(\overline{DE}\cong\overline{FG}\), \(\overline{EF}\cong\overline{GD}\), and \(\overline{DF}\cong\overline{DF}\), by the Side - Side - Side (SSS) congruence criterion, \(\triangle DEF\cong\triangle FGD\). So the reason for \(\triangle DEF\cong\triangle FGD\) is "SSS Congruence Theorem".

Answer:

1. \(FG\)
2. Given
3. \(DF\)
4. SSS Congruence Theorem