Question

given: (overline{eg}paralleloverline{df},overline{eg}congoverline{df}) prove: (overline{ed}paralleloverline{gf}) proof: 1. (overline{eg}congoverline{df}) 1. given 2. (egparallel df) 2. given 3. (angle egdcongangle fdg) 3. alt. int. (\triangle) thm. 4. (gdcong gd) 4. reflex. prop. of (cong) 5. (\triangle egdcong\triangle fdg) 5. sas steps 1, 3, and 4 6. (angle edgcongangle fgd) 6.

Explanation:

Step1: Analyze the given information

We are given that \(EG\parallel DF\) (so we can use the Alternate - Interior Angles Theorem for \(\angle EGD\) and \(\angle FDG\)), \(EG = DF\) (a side - equality), and \(GD=GD\) (reflexive property of equality).

Step2: Recall the SAS (Side - Angle - Side) congruence criterion

The SAS congruence criterion states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
In \(\triangle EGD\) and \(\triangle FDG\):

• \(EG = DF\) (given side)
• \(\angle EGD=\angle FDG\) (by the Alternate - Interior Angles Theorem since \(EG\parallel DF\))
• \(GD = GD\) (reflexive property of equality, common side)

Answer:

We use the SAS (Side - Angle - Side) congruence criterion to prove that \(\triangle EGD\cong\triangle FDG\). So, the reason is SAS (Side - Angle - Side) congruence.