Final answer:
Reflect ΔECD over . This establishes that ∠ACB ≅ ∠E'C'D'. Then, translate point D' to point B. This establishes that ∠E'D'C' ≅ ∠ABC. Therefore, ΔACB ~ ΔECD by the AA similarity postulate.
The answer is option 3) ∠ACB ≅ ∠E'C'D'; translate point D' to point B
Step-by-step explanation:
1. We are given that ∠ACB and ∠E'C'D' are congruent.
This means that the angle formed by points A, C, and B is equal in measure to the angle formed by points E', C', and D'.
2. To establish this congruence, we can perform a translation of point D' to point B.
A translation is a rigid transformation that moves all points in a figure the same distance in the same direction.
By translating point D' to point B, we ensure that the corresponding points in the two triangles coincide.
3. As a result of this translation, we can conclude that ∠ACB and ∠E'C'D' are congruent.
The translation preserves the shape, size, and orientation of the triangles, maintaining the equality of the corresponding angles.
The answer is option 3) ∠ACB ≅ ∠E'C'D'; translate point D' to point B
Your question is incomplete, but most probably the full question was:
Which of the following completes the proof?
Given: Segment AC is perpendicular to segment BD
Prove: ΔACB ~ ΔECD
Reflect ΔECD over . This establishes that ________. Then, ________. This establishes that ∠E'D'C' ≅ ∠ABC. Therefore, ΔACB ~ ΔECD by the AA similarity postulate.
Question 3 options:
- ∠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