Drag and drop the answers to the boxes to correctly complete the proof. Given: DEFG is a parallelogram.

Parallelogram D E F G with diagonal D F. Angle G D F is labeled 1, angle F D E is labeled 2, angle E F D is labeled 3, and angle D F G is labeled 4.
Prove: DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ .

Opposite sides of a parallelogram are parallel, so Response area because they are alternate interior angles. By the reflexive property of congruence, DF¯¯¯¯¯≅DF¯¯¯¯¯ . Therefore, △DGF≅△FED by the Response area . That means DE¯¯¯¯¯≅GF¯¯¯¯¯ and DG¯¯¯¯¯¯≅EF¯¯¯¯¯ by Response area.

and by CPCTC

Step-by-step explanation:

The given information are;

The shape of the given polygon DEFG = Parallelogram with diagonal DF

Therefore, we have;

Response area

and   (Opposite sides of a parallelogram are parallel)

∠GFD ≅ ∠EFD (Alternate interior angles)

∠DFG ≅ ∠FDE (Alternate interior angles)

(By the reflexive property of congruence)

Therefore

ΔDGF ≅ ΔFED (Angle-Side-Angle (ASA) condition of congruency)

Which gives;

and (Congruent Parts of Congruent Triangles are Congruent (CPCTC).

a 270° rotation about point p says on e d g e n u it y (had to space it out because it says it's a misspelling so i couldn't submit it