Read the proof. Given: AEEC; BDDC

Prove: △AEC ~ △BDC

Triangle A E C is shown. Line segment B D is drawn near point C to form triangle B D C.

Statement Reason
1. AEEC;BDDC 1. given
2. ∠AEC is a rt. ∠; ∠BDC is a rt. ∠ 2. definition of perpendicular
3. ∠AEC ≅ ∠BDC 3. all right angles are congruent
4. ? 4. reflexive property
5. △AEC ~ △BDC 5. AA similarity theorem
What is the missing statement in step 4?

∠ACE ≅ ∠BCD
∠EAB ≅ ∠DBC
∠EAC ≅ ∠EAC
∠CBD ≅ ∠DBC

table 3 represents a direct variation function. : )

i am not sure :

step-by-step explanation:

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

im guessing (23,14). hope this : )