The proof that δabc ≅ δcda is shown.

given: ∥ and ∥
prove: δabc ≅ δcda

what is the missing reason in the proof?

statements reasons
1. ab ∥ cd; bc ∥ da 1. given
2. quadrilateral abcd is a ▱ 2. definition of parallelogram
3. ab ≅ cd; bc ≅ da 3. opposite sides of a parallelogram are ≅
4. ac ≅ ac 4. reflexive property
5. δabc ≅ δcda 5. ?

a. perpendicular bisector theorem

b. pythagorean theorem

c. hl theorem

d. sss congruence theorem

D. SSS congruence theorem

Step-by-step explanation:

In the given proof we have in ΔABC ≅ ΔCDA

AB ≅ CD; BC ≅ DA[ opposite sides of a parallelogram are ≅]

AC ≅ AC[ reflexive property]

Therefore by SSS postulate of congruence  ΔABC ≅ ΔCDA.

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)

the answer is ! ?