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.

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step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.