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).

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

can i see the following ?

step-by-step explanation: