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 . Then, . This establishes that ∠E'D'C' ≅ ∠ABC. Therefore, ΔACB ~ ΔECD by the AA similarity postulate.

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

Therefore, the answer is the third option:∠ACB ≅ ∠E'C'D'; translate point D' to point B

Explanation:

1. First blank: ∠ACB ≅ ∠E'C'D'

Since segment AC is perpendicular to segment BD (given) and the point C is their intersection point, when you reflect triangle ECD over the segment AC, you get:

Hence: ∠ACB ≅ ∠E'C'D'

So far, you have established one pair of congruent angles.

2. Second blank: translate point D' to point B

You need to establish that other pair of angles are congruent.

Then, translate the triangle D'C'E' moving point D' to point B, which will show that angles ABC and E'D'C' are congruents.

Hence, you have proved a second pair of angles are congruent.

The AA (angle-angle) similarity postulate assures that two angles are similar if two pair of angles are congruent (because the third pair has to be congruent necessarily).

8 2/3 is the answer

step-by-step explanation: just use a calculator

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