Question

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, translate point E′ to point A. This establishes that ________. Therefore, ΔACB ~ ΔECD by the AA similarity postulate.

∠ABC ≅ ∠E′D′C′; ∠E′D′C′ ≅ ∠ABC
∠ACB ≅ ∠E′C′D′; ∠D′E′C′ ≅ ∠BAC
∠ACB ≅ ∠E′C′D′; ∠E′D′C′ ≅ ∠ABC
∠ABC ≅ ∠E′D′C′; ∠D′E′C′ ≅ ∠BAC

Answer 1

Reflecting ΔECD establishes that ∠ACB ≅ ∠E'C'D', and translating point E' to point A shows ∠E'D'C' ≅ ∠ABC, proving that ΔACB ~ ΔECD by the AA postulate.

Step-by-step explanation:

To complete the proof, you need to identify corresponding angles that are congruent after reflection and translation transformations. Reflecting ΔECD over segment AC would produce a triangle, let's call it ΔE'D'C', where ∠ACB is congruent to ∠E'C'D' because reflection preserves angle measures, and they are corresponding angles over the line of reflection. Then, by translating point E' to point A, the angle ∠E'D'C' would align with ∠ABC again preserving the angle measures. Therefore, we can conclude that ∠ACB ≅ ∠E'C'D' and ∠E'D'C' ≅ ∠ABC. This confirms the triangles ΔACB and ΔECD are similar by the AA (Angle-Angle) Similarity Postulate.