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).

4

step-by-step explanation:

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere