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.

A. translate point E' to point A; ∠E'D'C' ≅ ∠ABC
B. translate point D' to point A; ∠E'D'C' ≅ ∠BAC
C. translate point D' to point A; ∠D'E'C' ≅ ∠BAC
D. translate point E' to point A; ∠D'E'C ≅ ∠BAC

I think it's either A or D, but I feel like it could be either.

The answer is D. translate point E' to point A; ∠D'E'C' ≅ ∠BAC

Step-by-step explanation: Because you want to find another equal angle so when you translate point E' to point A both E' and A have to be in the middle because that is how angles work which means it can't be answer A cause neither E' or A are position in the middle of the angle where they are supposed to be.

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

the answers c