Which of the following completes the proof? Triangles ABC and EDC are formed from segments BD and AC, in which point C is between points B and D and point E is between points A and C.

Given: Segment AC is perpendicular to segment BD

Prove: ΔACB ~ ΔECD

Reflect ΔECD over segment AC. This establishes that ∠ACB ≅ ∠E'C'D'. Then, . This establishes that . Therefore, ΔACB ~ ΔECD by the AA similarity postulate.
translate point E' to point A; ∠E'D'C' ≅ ∠ABC
translate point D' to point A; ∠E'D'C' ≅ ∠BAC
translate point D' to point A; ∠D'E'C' ≅ ∠BAC
translate point E' to point A; ∠D'E'C' ≅ ∠BAC

From the given question, the required answer is option D. It can be deduced that: Reflect ΔECD over segment AC. This establishes that ∠ACB ≅ ∠E'C'D'. Then, (translate point E' to point A). This establishes that (∠D'E'C' ≅ ∠BAC). Therefore, ΔACB ~ ΔECD by the AA similarity postulate.

Rigid transformation is a process of resizing a given shape or figure. This can be done by either increasing or decreasing its dimensions (size) or changing its orientation. Some of the methods required in rigid transformation are: reflection, translation, rotation etc.

From the given question,

if ΔECD is first reflected over (turned about) segment AC, it would be deduced that m∠ACB ≅ m∠E'C'D' = (right angle principle).

Then, the next process is to translate (extend) point E' to point A, so that m∠D'E'C' ≅ m∠BAC.

This would also ensure that m∠E'D'C' ≅ m∠ABC due to the translation.

This implies that ΔACB ~ ΔECD by the AA similarity postulate.

Therefore, performing reflection and translation operations on ΔECD step-wisely can be used to prove that ΔACB ~ ΔECD by the AA similarity postulate.

A sketch for more clarifications is herewith attached to this answer.

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x equals 1

step-by-step explanation:

f(4)=f(×)

x=1

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)