Question

Drag and drop the answers to the boxes to correctly complete the proof.

Answer 1

Answer:

$\overline{DE} \cong \overline{GF}$ and $\overline{DG} \cong \overline{EF}$ 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

$\overline{DE} \left | \right | \overline{GF}$ and  $\overline{DG} \left | \right | \overline{EF}$ (Opposite sides of a parallelogram are parallel)

∠GFD ≅ ∠EFD (Alternate interior angles)

∠DFG ≅ ∠FDE (Alternate interior angles)

$\overline{DF} \cong \overline{DF}$ (By the reflexive property of congruence)

Therefore

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

Which gives;

$\overline{DE} \cong \overline{GF}$ and $\overline{DG} \cong \overline{EF}$ (Congruent Parts of Congruent Triangles are Congruent (CPCTC).